/*
 * Copyright (c) 2009, 2019, Oracle and/or its affiliates. All rights reserved.
 * ORACLE PROPRIETARY/CONFIDENTIAL. Use is subject to license terms.
 *
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 *
 */

package java.util;

import java.util.concurrent.CountedCompleter;
import java.util.concurrent.RecursiveTask;

/**
 * This class implements powerful and fully optimized versions, both
 * sequential and parallel, of the Dual-Pivot Quicksort algorithm by
 * Vladimir Yaroslavskiy, Jon Bentley and Josh Bloch. This algorithm
 * offers O(n log(n)) performance on all data sets, and is typically
 * faster than traditional (one-pivot) Quicksort implementations.
 *
 * There are also additional algorithms, invoked from the Dual-Pivot
 * Quicksort, such as mixed insertion sort, merging of runs and heap
 * sort, counting sort and parallel merge sort.
 *
 * @author Vladimir Yaroslavskiy
 * @author Jon Bentley
 * @author Josh Bloch
 * @author Doug Lea
 *
 * @version 2018.08.18
 *
 * @since 1.7 * 14
 */
final class DualPivotQuicksort {

    /**
     * Prevents instantiation.
     */
    private DualPivotQuicksort() {}

    /**
     * Max array size to use mixed insertion sort.
     */
    private static final int MAX_MIXED_INSERTION_SORT_SIZE = 65;

    /**
     * Max array size to use insertion sort.
     */
    private static final int MAX_INSERTION_SORT_SIZE = 44;

    /**
     * Min array size to perform sorting in parallel.
     */
    private static final int MIN_PARALLEL_SORT_SIZE = 4 << 10;

    /**
     * Min array size to try merging of runs.
     */
    private static final int MIN_TRY_MERGE_SIZE = 4 << 10;

    /**
     * Min size of the first run to continue with scanning.
     */
    private static final int MIN_FIRST_RUN_SIZE = 16;

    /**
     * Min factor for the first runs to continue scanning.
     */
    private static final int MIN_FIRST_RUNS_FACTOR = 7;

    /**
     * Max capacity of the index array for tracking runs.
     */
    private static final int MAX_RUN_CAPACITY = 5 << 10;

    /**
     * Min number of runs, required by parallel merging.
     */
    private static final int MIN_RUN_COUNT = 4;

    /**
     * Min array size to use parallel merging of parts.
     */
    private static final int MIN_PARALLEL_MERGE_PARTS_SIZE = 4 << 10;

    /**
     * Min size of a byte array to use counting sort.
     */
    private static final int MIN_BYTE_COUNTING_SORT_SIZE = 64;

    /**
     * Min size of a short or char array to use counting sort.
     */
    private static final int MIN_SHORT_OR_CHAR_COUNTING_SORT_SIZE = 1750;

    /**
     * Threshold of mixed insertion sort is incremented by this value.
     */
    private static final int DELTA = 3 << 1;

    /**
     * Max recursive partitioning depth before using heap sort.
     */
    private static final int MAX_RECURSION_DEPTH = 64 * DELTA;

    /**
     * Calculates the double depth of parallel merging.
     * Depth is negative, if tasks split before sorting.
     *
     * @param parallelism the parallelism level
     * @param size the target size
     * @return the depth of parallel merging
     */
    private static int getDepth(int parallelism, int size) {
        int depth = 0;

        while ((parallelism >>= 3) > 0 && (size >>= 2) > 0) {
            depth -= 2;
        }
        return depth;
    }

    /**
     * Sorts the specified range of the array using parallel merge
     * sort and/or Dual-Pivot Quicksort.
     *
     * To balance the faster splitting and parallelism of merge sort
     * with the faster element partitioning of Quicksort, ranges are
     * subdivided in tiers such that, if there is enough parallelism,
     * the four-way parallel merge is started, still ensuring enough
     * parallelism to process the partitions.
     *
     * @param a the array to be sorted
     * @param parallelism the parallelism level
     * @param low the index of the first element, inclusive, to be sorted
     * @param high the index of the last element, exclusive, to be sorted
     */
    static void sort(int[] a, int parallelism, int low, int high) {
        int size = high - low;

        if (parallelism > 1 && size > MIN_PARALLEL_SORT_SIZE) {
            int depth = getDepth(parallelism, size >> 12);
            int[] b = depth == 0 ? null : new int[size];
            new Sorter(null, a, b, low, size, low, depth).invoke();
        } else {
            sort(null, a, 0, low, high);
        }
    }

    /**
     * Sorts the specified array using the Dual-Pivot Quicksort and/or
     * other sorts in special-cases, possibly with parallel partitions.
     *
     * @param sorter parallel context
     * @param a the array to be sorted
     * @param bits the combination of recursion depth and bit flag, where
     *        the right bit "0" indicates that array is the leftmost part
     * @param low the index of the first element, inclusive, to be sorted
     * @param high the index of the last element, exclusive, to be sorted
     */
    static void sort(Sorter sorter, int[] a, int bits, int low, int high) {
        while (true) {
            int end = high - 1, size = high - low;

            /*
             * Run mixed insertion sort on small non-leftmost parts.
             */
            if (size < MAX_MIXED_INSERTION_SORT_SIZE + bits && (bits & 1) > 0) {
                mixedInsertionSort(a, low, high - 3 * ((size >> 5) << 3), high);
                return;
            }

            /*
             * Invoke insertion sort on small leftmost part.
             */
            if (size < MAX_INSERTION_SORT_SIZE) {
                insertionSort(a, low, high);
                return;
            }

            /*
             * Check if the whole array or large non-leftmost
             * parts are nearly sorted and then merge runs.
             */
            if ((bits == 0 || size > MIN_TRY_MERGE_SIZE && (bits & 1) > 0)
                    && tryMergeRuns(sorter, a, low, size)) {
                return;
            }

            /*
             * Switch to heap sort if execution
             * time is becoming quadratic.
             */
            if ((bits += DELTA) > MAX_RECURSION_DEPTH) {
                heapSort(a, low, high);
                return;
            }

            /*
             * Use an inexpensive approximation of the golden ratio
             * to select five sample elements and determine pivots.
             */
            int step = (size >> 3) * 3 + 3;

            /*
             * Five elements around (and including) the central element
             * will be used for pivot selection as described below. The
             * unequal choice of spacing these elements was empirically
             * determined to work well on a wide variety of inputs.
             */
            int e1 = low + step;
            int e5 = end - step;
            int e3 = (e1 + e5) >>> 1;
            int e2 = (e1 + e3) >>> 1;
            int e4 = (e3 + e5) >>> 1;
            int a3 = a[e3];

            /*
             * Sort these elements in place by the combination
             * of 4-element sorting network and insertion sort.
             *
             *    5 ------o-----------o------------
             *            |           |
             *    4 ------|-----o-----o-----o------
             *            |     |           |
             *    2 ------o-----|-----o-----o------
             *                  |     |
             *    1 ------------o-----o------------
             */
            if (a[e5] < a[e2]) { int t = a[e5]; a[e5] = a[e2]; a[e2] = t; }
            if (a[e4] < a[e1]) { int t = a[e4]; a[e4] = a[e1]; a[e1] = t; }
            if (a[e5] < a[e4]) { int t = a[e5]; a[e5] = a[e4]; a[e4] = t; }
            if (a[e2] < a[e1]) { int t = a[e2]; a[e2] = a[e1]; a[e1] = t; }
            if (a[e4] < a[e2]) { int t = a[e4]; a[e4] = a[e2]; a[e2] = t; }

            if (a3 < a[e2]) {
                if (a3 < a[e1]) {
                    a[e3] = a[e2]; a[e2] = a[e1]; a[e1] = a3;
                } else {
                    a[e3] = a[e2]; a[e2] = a3;
                }
            } else if (a3 > a[e4]) {
                if (a3 > a[e5]) {
                    a[e3] = a[e4]; a[e4] = a[e5]; a[e5] = a3;
                } else {
                    a[e3] = a[e4]; a[e4] = a3;
                }
            }

            // Pointers
            int lower = low; // The index of the last element of the left part
            int upper = end; // The index of the first element of the right part

            /*
             * Partitioning with 2 pivots in case of different elements.
             */
            if (a[e1] < a[e2] && a[e2] < a[e3] && a[e3] < a[e4] && a[e4] < a[e5]) {

                /*
                 * Use the first and fifth of the five sorted elements as
                 * the pivots. These values are inexpensive approximation
                 * of tertiles. Note, that pivot1 < pivot2.
                 */
                int pivot1 = a[e1];
                int pivot2 = a[e5];

                /*
                 * The first and the last elements to be sorted are moved
                 * to the locations formerly occupied by the pivots. When
                 * partitioning is completed, the pivots are swapped back
                 * into their final positions, and excluded from the next
                 * subsequent sorting.
                 */
                a[e1] = a[lower];
                a[e5] = a[upper];

                /*
                 * Skip elements, which are less or greater than the pivots.
                 */
                while (a[++lower] < pivot1);
                while (a[--upper] > pivot2);

                /*
                 * Backward 3-interval partitioning
                 *
                 *   left part                 central part          right part
                 * +------------------------------------------------------------+
                 * |  < pivot1  |   ?   |  pivot1 <= && <= pivot2  |  > pivot2  |
                 * +------------------------------------------------------------+
                 *             ^       ^                            ^
                 *             |       |                            |
                 *           lower     k                          upper
                 *
                 * Invariants:
                 *
                 *              all in (low, lower] < pivot1
                 *    pivot1 <= all in (k, upper)  <= pivot2
                 *              all in [upper, end) > pivot2
                 *
                 * Pointer k is the last index of ?-part
                 */
                for (int unused = --lower, k = ++upper; --k > lower; ) {
                    int ak = a[k];

                    if (ak < pivot1) { // Move a[k] to the left side
                        while (lower < k) {
                            if (a[++lower] >= pivot1) {
                                if (a[lower] > pivot2) {
                                    a[k] = a[--upper];
                                    a[upper] = a[lower];
                                } else {
                                    a[k] = a[lower];
                                }
                                a[lower] = ak;
                                break;
                            }
                        }
                    } else if (ak > pivot2) { // Move a[k] to the right side
                        a[k] = a[--upper];
                        a[upper] = ak;
                    }
                }

                /*
                 * Swap the pivots into their final positions.
                 */
                a[low] = a[lower]; a[lower] = pivot1;
                a[end] = a[upper]; a[upper] = pivot2;

                /*
                 * Sort non-left parts recursively (possibly in parallel),
                 * excluding known pivots.
                 */
                if (size > MIN_PARALLEL_SORT_SIZE && sorter != null) {
                    sorter.forkSorter(bits | 1, lower + 1, upper);
                    sorter.forkSorter(bits | 1, upper + 1, high);
                } else {
                    sort(sorter, a, bits | 1, lower + 1, upper);
                    sort(sorter, a, bits | 1, upper + 1, high);
                }

            } else { // Use single pivot in case of many equal elements

                /*
                 * Use the third of the five sorted elements as the pivot.
                 * This value is inexpensive approximation of the median.
                 */
                int pivot = a[e3];

                /*
                 * The first element to be sorted is moved to the
                 * location formerly occupied by the pivot. After
                 * completion of partitioning the pivot is swapped
                 * back into its final position, and excluded from
                 * the next subsequent sorting.
                 */
                a[e3] = a[lower];

                /*
                 * Traditional 3-way (Dutch National Flag) partitioning
                 *
                 *   left part                 central part    right part
                 * +------------------------------------------------------+
                 * |   < pivot   |     ?     |   == pivot   |   > pivot   |
                 * +------------------------------------------------------+
                 *              ^           ^                ^
                 *              |           |                |
                 *            lower         k              upper
                 *
                 * Invariants:
                 *
                 *   all in (low, lower] < pivot
                 *   all in (k, upper)  == pivot
                 *   all in [upper, end] > pivot
                 *
                 * Pointer k is the last index of ?-part
                 */
                for (int k = ++upper; --k > lower; ) {
                    int ak = a[k];

                    if (ak != pivot) {
                        a[k] = pivot;

                        if (ak < pivot) { // Move a[k] to the left side
                            while (a[++lower] < pivot);

                            if (a[lower] > pivot) {
                                a[--upper] = a[lower];
                            }
                            a[lower] = ak;
                        } else { // ak > pivot - Move a[k] to the right side
                            a[--upper] = ak;
                        }
                    }
                }

                /*
                 * Swap the pivot into its final position.
                 */
                a[low] = a[lower]; a[lower] = pivot;

                /*
                 * Sort the right part (possibly in parallel), excluding
                 * known pivot. All elements from the central part are
                 * equal and therefore already sorted.
                 */
                if (size > MIN_PARALLEL_SORT_SIZE && sorter != null) {
                    sorter.forkSorter(bits | 1, upper, high);
                } else {
                    sort(sorter, a, bits | 1, upper, high);
                }
            }
            high = lower; // Iterate along the left part
        }
    }

    /**
     * Sorts the specified range of the array using mixed insertion sort.
     *
     * Mixed insertion sort is combination of simple insertion sort,
     * pin insertion sort and pair insertion sort.
     *
     * In the context of Dual-Pivot Quicksort, the pivot element
     * from the left part plays the role of sentinel, because it
     * is less than any elements from the given part. Therefore,
     * expensive check of the left range can be skipped on each
     * iteration unless it is the leftmost call.
     *
     * @param a the array to be sorted
     * @param low the index of the first element, inclusive, to be sorted
     * @param end the index of the last element for simple insertion sort
     * @param high the index of the last element, exclusive, to be sorted
     */
    private static void mixedInsertionSort(int[] a, int low, int end, int high) {
        if (end == high) {

            /*
             * Invoke simple insertion sort on tiny array.
             */
            for (int i; ++low < end; ) {
                int ai = a[i = low];

                while (ai < a[--i]) {
                    a[i + 1] = a[i];
                }
                a[i + 1] = ai;
            }
        } else {

            /*
             * Start with pin insertion sort on small part.
             *
             * Pin insertion sort is extended simple insertion sort.
             * The main idea of this sort is to put elements larger
             * than an element called pin to the end of array (the
             * proper area for such elements). It avoids expensive
             * movements of these elements through the whole array.
             */
            int pin = a[end];

            for (int i, p = high; ++low < end; ) {
                int ai = a[i = low];

                if (ai < a[i - 1]) { // Small element

                    /*
                     * Insert small element into sorted part.
                     */
                    a[i] = a[--i];

                    while (ai < a[--i]) {
                        a[i + 1] = a[i];
                    }
                    a[i + 1] = ai;

                } else if (p > i && ai > pin) { // Large element

                    /*
                     * Find element smaller than pin.
                     */
                    while (a[--p] > pin);

                    /*
                     * Swap it with large element.
                     */
                    if (p > i) {
                        ai = a[p];
                        a[p] = a[i];
                    }

                    /*
                     * Insert small element into sorted part.
                     */
                    while (ai < a[--i]) {
                        a[i + 1] = a[i];
                    }
                    a[i + 1] = ai;
                }
            }

            /*
             * Continue with pair insertion sort on remain part.
             */
            for (int i; low < high; ++low) {
                int a1 = a[i = low], a2 = a[++low];

                /*
                 * Insert two elements per iteration: at first, insert the
                 * larger element and then insert the smaller element, but
                 * from the position where the larger element was inserted.
                 */
                if (a1 > a2) {

                    while (a1 < a[--i]) {
                        a[i + 2] = a[i];
                    }
                    a[++i + 1] = a1;

                    while (a2 < a[--i]) {
                        a[i + 1] = a[i];
                    }
                    a[i + 1] = a2;

                } else if (a1 < a[i - 1]) {

                    while (a2 < a[--i]) {
                        a[i + 2] = a[i];
                    }
                    a[++i + 1] = a2;

                    while (a1 < a[--i]) {
                        a[i + 1] = a[i];
                    }
                    a[i + 1] = a1;
                }
            }
        }
    }

    /**
     * Sorts the specified range of the array using insertion sort.
     *
     * @param a the array to be sorted
     * @param low the index of the first element, inclusive, to be sorted
     * @param high the index of the last element, exclusive, to be sorted
     */
    private static void insertionSort(int[] a, int low, int high) {
        for (int i, k = low; ++k < high; ) {
            int ai = a[i = k];

            if (ai < a[i - 1]) {
                while (--i >= low && ai < a[i]) {
                    a[i + 1] = a[i];
                }
                a[i + 1] = ai;
            }
        }
    }

    /**
     * Sorts the specified range of the array using heap sort.
     *
     * @param a the array to be sorted
     * @param low the index of the first element, inclusive, to be sorted
     * @param high the index of the last element, exclusive, to be sorted
     */
    private static void heapSort(int[] a, int low, int high) {
        for (int k = (low + high) >>> 1; k > low; ) {
            pushDown(a, --k, a[k], low, high);
        }
        while (--high > low) {
            int max = a[low];
            pushDown(a, low, a[high], low, high);
            a[high] = max;
        }
    }

    /**
     * Pushes specified element down during heap sort.
     *
     * @param a the given array
     * @param p the start index
     * @param value the given element
     * @param low the index of the first element, inclusive, to be sorted
     * @param high the index of the last element, exclusive, to be sorted
     */
    private static void pushDown(int[] a, int p, int value, int low, int high) {
        for (int k ;; a[p] = a[p = k]) {
            k = (p << 1) - low + 2; // Index of the right child

            if (k > high) {
                break;
            }
            if (k == high || a[k] < a[k - 1]) {
                --k;
            }
            if (a[k] <= value) {
                break;
            }
        }
        a[p] = value;
    }

    /**
     * Tries to sort the specified range of the array.
     *
     * @param sorter parallel context
     * @param a the array to be sorted
     * @param low the index of the first element to be sorted
     * @param size the array size
     * @return true if finally sorted, false otherwise
     */
    private static boolean tryMergeRuns(Sorter sorter, int[] a, int low, int size) {

        /*
         * The run array is constructed only if initial runs are
         * long enough to continue, run[i] then holds start index
         * of the i-th sequence of elements in non-descending order.
         */
        int[] run = null;
        int high = low + size;
        int count = 1, last = low;

        /*
         * Identify all possible runs.
         */
        for (int k = low + 1; k < high; ) {

            /*
             * Find the end index of the current run.
             */
            if (a[k - 1] < a[k]) {

                // Identify ascending sequence
                while (++k < high && a[k - 1] <= a[k]);

            } else if (a[k - 1] > a[k]) {

                // Identify descending sequence
                while (++k < high && a[k - 1] >= a[k]);

                // Reverse into ascending order
                for (int i = last - 1, j = k; ++i < --j && a[i] > a[j]; ) {
                    int ai = a[i]; a[i] = a[j]; a[j] = ai;
                }
            } else { // Identify constant sequence
                for (int ak = a[k]; ++k < high && ak == a[k]; );

                if (k < high) {
                    continue;
                }
            }

            /*
             * Check special cases.
             */
            if (run == null) {
                if (k == high) {

                    /*
                     * The array is monotonous sequence,
                     * and therefore already sorted.
                     */
                    return true;
                }

                if (k - low < MIN_FIRST_RUN_SIZE) {

                    /*
                     * The first run is too small
                     * to proceed with scanning.
                     */
                    return false;
                }

                run = new int[((size >> 10) | 0x7F) & 0x3FF];
                run[0] = low;

            } else if (a[last - 1] > a[last]) {

                if (count > (k - low) >> MIN_FIRST_RUNS_FACTOR) {

                    /*
                     * The first runs are not long
                     * enough to continue scanning.
                     */
                    return false;
                }

                if (++count == MAX_RUN_CAPACITY) {

                    /*
                     * Array is not highly structured.
                     */
                    return false;
                }

                if (count == run.length) {

                    /*
                     * Increase capacity of index array.
                     */
                    run = Arrays.copyOf(run, count << 1);
                }
            }
            run[count] = (last = k);
        }

        /*
         * Merge runs of highly structured array.
         */
        if (count > 1) {
            int[] b; int offset = low;

            if (sorter == null || (b = (int[]) sorter.b) == null) {
                b = new int[size];
            } else {
                offset = sorter.offset;
            }
            mergeRuns(a, b, offset, 1, sorter != null, run, 0, count);
        }
        return true;
    }

    /**
     * Merges the specified runs.
     *
     * @param a the source array
     * @param b the temporary buffer used in merging
     * @param offset the start index in the source, inclusive
     * @param aim specifies merging: to source ( > 0), buffer ( < 0) or any ( == 0)
     * @param parallel indicates whether merging is performed in parallel
     * @param run the start indexes of the runs, inclusive
     * @param lo the start index of the first run, inclusive
     * @param hi the start index of the last run, inclusive
     * @return the destination where runs are merged
     */
    private static int[] mergeRuns(int[] a, int[] b, int offset,
            int aim, boolean parallel, int[] run, int lo, int hi) {

        if (hi - lo == 1) {
            if (aim >= 0) {
                return a;
            }
            for (int i = run[hi], j = i - offset, low = run[lo]; i > low;
                b[--j] = a[--i]
            );
            return b;
        }

        /*
         * Split into approximately equal parts.
         */
        int mi = lo, rmi = (run[lo] + run[hi]) >>> 1;
        while (run[++mi + 1] <= rmi);

        /*
         * Merge the left and right parts.
         */
        int[] a1, a2;

        if (parallel && hi - lo > MIN_RUN_COUNT) {
            RunMerger merger = new RunMerger(a, b, offset, 0, run, mi, hi).forkMe();
            a1 = mergeRuns(a, b, offset, -aim, true, run, lo, mi);
            a2 = (int[]) merger.getDestination();
        } else {
            a1 = mergeRuns(a, b, offset, -aim, false, run, lo, mi);
            a2 = mergeRuns(a, b, offset,    0, false, run, mi, hi);
        }

        int[] dst = a1 == a ? b : a;

        int k   = a1 == a ? run[lo] - offset : run[lo];
        int lo1 = a1 == b ? run[lo] - offset : run[lo];
        int hi1 = a1 == b ? run[mi] - offset : run[mi];
        int lo2 = a2 == b ? run[mi] - offset : run[mi];
        int hi2 = a2 == b ? run[hi] - offset : run[hi];

        if (parallel) {
            new Merger(null, dst, k, a1, lo1, hi1, a2, lo2, hi2).invoke();
        } else {
            mergeParts(null, dst, k, a1, lo1, hi1, a2, lo2, hi2);
        }
        return dst;
    }

    /**
     * Merges the sorted parts.
     *
     * @param merger parallel context
     * @param dst the destination where parts are merged
     * @param k the start index of the destination, inclusive
     * @param a1 the first part
     * @param lo1 the start index of the first part, inclusive
     * @param hi1 the end index of the first part, exclusive
     * @param a2 the second part
     * @param lo2 the start index of the second part, inclusive
     * @param hi2 the end index of the second part, exclusive
     */
    private static void mergeParts(Merger merger, int[] dst, int k,
            int[] a1, int lo1, int hi1, int[] a2, int lo2, int hi2) {

        if (merger != null && a1 == a2) {

            while (true) {

                /*
                 * The first part must be larger.
                 */
                if (hi1 - lo1 < hi2 - lo2) {
                    int lo = lo1; lo1 = lo2; lo2 = lo;
                    int hi = hi1; hi1 = hi2; hi2 = hi;
                }

                /*
                 * Small parts will be merged sequentially.
                 */
                if (hi1 - lo1 < MIN_PARALLEL_MERGE_PARTS_SIZE) {
                    break;
                }

                /*
                 * Find the median of the larger part.
                 */
                int mi1 = (lo1 + hi1) >>> 1;
                int key = a1[mi1];
                int mi2 = hi2;

                /*
                 * Partition the smaller part.
                 */
                for (int loo = lo2; loo < mi2; ) {
                    int t = (loo + mi2) >>> 1;

                    if (key > a2[t]) {
                        loo = t + 1;
                    } else {
                        mi2 = t;
                    }
                }

                int d = mi2 - lo2 + mi1 - lo1;

                /*
                 * Merge the right sub-parts in parallel.
                 */
                merger.forkMerger(dst, k + d, a1, mi1, hi1, a2, mi2, hi2);

                /*
                 * Process the sub-left parts.
                 */
                hi1 = mi1;
                hi2 = mi2;
            }
        }

        /*
         * Merge small parts sequentially.
         */
        while (lo1 < hi1 && lo2 < hi2) {
            dst[k++] = a1[lo1] < a2[lo2] ? a1[lo1++] : a2[lo2++];
        }
        if (dst != a1 || k < lo1) {
            while (lo1 < hi1) {
                dst[k++] = a1[lo1++];
            }
        }
        if (dst != a2 || k < lo2) {
            while (lo2 < hi2) {
                dst[k++] = a2[lo2++];
            }
        }
    }

// [long]

    /**
     * Sorts the specified range of the array using parallel merge
     * sort and/or Dual-Pivot Quicksort.
     *
     * To balance the faster splitting and parallelism of merge sort
     * with the faster element partitioning of Quicksort, ranges are
     * subdivided in tiers such that, if there is enough parallelism,
     * the four-way parallel merge is started, still ensuring enough
     * parallelism to process the partitions.
     *
     * @param a the array to be sorted
     * @param parallelism the parallelism level
     * @param low the index of the first element, inclusive, to be sorted
     * @param high the index of the last element, exclusive, to be sorted
     */
    static void sort(long[] a, int parallelism, int low, int high) {
        int size = high - low;

        if (parallelism > 1 && size > MIN_PARALLEL_SORT_SIZE) {
            int depth = getDepth(parallelism, size >> 12);
            long[] b = depth == 0 ? null : new long[size];
            new Sorter(null, a, b, low, size, low, depth).invoke();
        } else {
            sort(null, a, 0, low, high);
        }
    }

    /**
     * Sorts the specified array using the Dual-Pivot Quicksort and/or
     * other sorts in special-cases, possibly with parallel partitions.
     *
     * @param sorter parallel context
     * @param a the array to be sorted
     * @param bits the combination of recursion depth and bit flag, where
     *        the right bit "0" indicates that array is the leftmost part
     * @param low the index of the first element, inclusive, to be sorted
     * @param high the index of the last element, exclusive, to be sorted
     */
    static void sort(Sorter sorter, long[] a, int bits, int low, int high) {
        while (true) {
            int end = high - 1, size = high - low;

            /*
             * Run mixed insertion sort on small non-leftmost parts.
             */
            if (size < MAX_MIXED_INSERTION_SORT_SIZE + bits && (bits & 1) > 0) {
                mixedInsertionSort(a, low, high - 3 * ((size >> 5) << 3), high);
                return;
            }

            /*
             * Invoke insertion sort on small leftmost part.
             */
            if (size < MAX_INSERTION_SORT_SIZE) {
                insertionSort(a, low, high);
                return;
            }

            /*
             * Check if the whole array or large non-leftmost
             * parts are nearly sorted and then merge runs.
             */
            if ((bits == 0 || size > MIN_TRY_MERGE_SIZE && (bits & 1) > 0)
                    && tryMergeRuns(sorter, a, low, size)) {
                return;
            }

            /*
             * Switch to heap sort if execution
             * time is becoming quadratic.
             */
            if ((bits += DELTA) > MAX_RECURSION_DEPTH) {
                heapSort(a, low, high);
                return;
            }

            /*
             * Use an inexpensive approximation of the golden ratio
             * to select five sample elements and determine pivots.
             */
            int step = (size >> 3) * 3 + 3;

            /*
             * Five elements around (and including) the central element
             * will be used for pivot selection as described below. The
             * unequal choice of spacing these elements was empirically
             * determined to work well on a wide variety of inputs.
             */
            int e1 = low + step;
            int e5 = end - step;
            int e3 = (e1 + e5) >>> 1;
            int e2 = (e1 + e3) >>> 1;
            int e4 = (e3 + e5) >>> 1;
            long a3 = a[e3];

            /*
             * Sort these elements in place by the combination
             * of 4-element sorting network and insertion sort.
             *
             *    5 ------o-----------o------------
             *            |           |
             *    4 ------|-----o-----o-----o------
             *            |     |           |
             *    2 ------o-----|-----o-----o------
             *                  |     |
             *    1 ------------o-----o------------
             */
            if (a[e5] < a[e2]) { long t = a[e5]; a[e5] = a[e2]; a[e2] = t; }
            if (a[e4] < a[e1]) { long t = a[e4]; a[e4] = a[e1]; a[e1] = t; }
            if (a[e5] < a[e4]) { long t = a[e5]; a[e5] = a[e4]; a[e4] = t; }
            if (a[e2] < a[e1]) { long t = a[e2]; a[e2] = a[e1]; a[e1] = t; }
            if (a[e4] < a[e2]) { long t = a[e4]; a[e4] = a[e2]; a[e2] = t; }

            if (a3 < a[e2]) {
                if (a3 < a[e1]) {
                    a[e3] = a[e2]; a[e2] = a[e1]; a[e1] = a3;
                } else {
                    a[e3] = a[e2]; a[e2] = a3;
                }
            } else if (a3 > a[e4]) {
                if (a3 > a[e5]) {
                    a[e3] = a[e4]; a[e4] = a[e5]; a[e5] = a3;
                } else {
                    a[e3] = a[e4]; a[e4] = a3;
                }
            }

            // Pointers
            int lower = low; // The index of the last element of the left part
            int upper = end; // The index of the first element of the right part

            /*
             * Partitioning with 2 pivots in case of different elements.
             */
            if (a[e1] < a[e2] && a[e2] < a[e3] && a[e3] < a[e4] && a[e4] < a[e5]) {

                /*
                 * Use the first and fifth of the five sorted elements as
                 * the pivots. These values are inexpensive approximation
                 * of tertiles. Note, that pivot1 < pivot2.
                 */
                long pivot1 = a[e1];
                long pivot2 = a[e5];

                /*
                 * The first and the last elements to be sorted are moved
                 * to the locations formerly occupied by the pivots. When
                 * partitioning is completed, the pivots are swapped back
                 * into their final positions, and excluded from the next
                 * subsequent sorting.
                 */
                a[e1] = a[lower];
                a[e5] = a[upper];

                /*
                 * Skip elements, which are less or greater than the pivots.
                 */
                while (a[++lower] < pivot1);
                while (a[--upper] > pivot2);

                /*
                 * Backward 3-interval partitioning
                 *
                 *   left part                 central part          right part
                 * +------------------------------------------------------------+
                 * |  < pivot1  |   ?   |  pivot1 <= && <= pivot2  |  > pivot2  |
                 * +------------------------------------------------------------+
                 *             ^       ^                            ^
                 *             |       |                            |
                 *           lower     k                          upper
                 *
                 * Invariants:
                 *
                 *              all in (low, lower] < pivot1
                 *    pivot1 <= all in (k, upper)  <= pivot2
                 *              all in [upper, end) > pivot2
                 *
                 * Pointer k is the last index of ?-part
                 */
                for (int unused = --lower, k = ++upper; --k > lower; ) {
                    long ak = a[k];

                    if (ak < pivot1) { // Move a[k] to the left side
                        while (lower < k) {
                            if (a[++lower] >= pivot1) {
                                if (a[lower] > pivot2) {
                                    a[k] = a[--upper];
                                    a[upper] = a[lower];
                                } else {
                                    a[k] = a[lower];
                                }
                                a[lower] = ak;
                                break;
                            }
                        }
                    } else if (ak > pivot2) { // Move a[k] to the right side
                        a[k] = a[--upper];
                        a[upper] = ak;
                    }
                }

                /*
                 * Swap the pivots into their final positions.
                 */
                a[low] = a[lower]; a[lower] = pivot1;
                a[end] = a[upper]; a[upper] = pivot2;

                /*
                 * Sort non-left parts recursively (possibly in parallel),
                 * excluding known pivots.
                 */
                if (size > MIN_PARALLEL_SORT_SIZE && sorter != null) {
                    sorter.forkSorter(bits | 1, lower + 1, upper);
                    sorter.forkSorter(bits | 1, upper + 1, high);
                } else {
                    sort(sorter, a, bits | 1, lower + 1, upper);
                    sort(sorter, a, bits | 1, upper + 1, high);
                }

            } else { // Use single pivot in case of many equal elements

                /*
                 * Use the third of the five sorted elements as the pivot.
                 * This value is inexpensive approximation of the median.
                 */
                long pivot = a[e3];

                /*
                 * The first element to be sorted is moved to the
                 * location formerly occupied by the pivot. After
                 * completion of partitioning the pivot is swapped
                 * back into its final position, and excluded from
                 * the next subsequent sorting.
                 */
                a[e3] = a[lower];

                /*
                 * Traditional 3-way (Dutch National Flag) partitioning
                 *
                 *   left part                 central part    right part
                 * +------------------------------------------------------+
                 * |   < pivot   |     ?     |   == pivot   |   > pivot   |
                 * +------------------------------------------------------+
                 *              ^           ^                ^
                 *              |           |                |
                 *            lower         k              upper
                 *
                 * Invariants:
                 *
                 *   all in (low, lower] < pivot
                 *   all in (k, upper)  == pivot
                 *   all in [upper, end] > pivot
                 *
                 * Pointer k is the last index of ?-part
                 */
                for (int k = ++upper; --k > lower; ) {
                    long ak = a[k];

                    if (ak != pivot) {
                        a[k] = pivot;

                        if (ak < pivot) { // Move a[k] to the left side
                            while (a[++lower] < pivot);

                            if (a[lower] > pivot) {
                                a[--upper] = a[lower];
                            }
                            a[lower] = ak;
                        } else { // ak > pivot - Move a[k] to the right side
                            a[--upper] = ak;
                        }
                    }
                }

                /*
                 * Swap the pivot into its final position.
                 */
                a[low] = a[lower]; a[lower] = pivot;

                /*
                 * Sort the right part (possibly in parallel), excluding
                 * known pivot. All elements from the central part are
                 * equal and therefore already sorted.
                 */
                if (size > MIN_PARALLEL_SORT_SIZE && sorter != null) {
                    sorter.forkSorter(bits | 1, upper, high);
                } else {
                    sort(sorter, a, bits | 1, upper, high);
                }
            }
            high = lower; // Iterate along the left part
        }
    }

    /**
     * Sorts the specified range of the array using mixed insertion sort.
     *
     * Mixed insertion sort is combination of simple insertion sort,
     * pin insertion sort and pair insertion sort.
     *
     * In the context of Dual-Pivot Quicksort, the pivot element
     * from the left part plays the role of sentinel, because it
     * is less than any elements from the given part. Therefore,
     * expensive check of the left range can be skipped on each
     * iteration unless it is the leftmost call.
     *
     * @param a the array to be sorted
     * @param low the index of the first element, inclusive, to be sorted
     * @param end the index of the last element for simple insertion sort
     * @param high the index of the last element, exclusive, to be sorted
     */
    private static void mixedInsertionSort(long[] a, int low, int end, int high) {
        if (end == high) {

            /*
             * Invoke simple insertion sort on tiny array.
             */
            for (int i; ++low < end; ) {
                long ai = a[i = low];

                while (ai < a[--i]) {
                    a[i + 1] = a[i];
                }
                a[i + 1] = ai;
            }
        } else {

            /*
             * Start with pin insertion sort on small part.
             *
             * Pin insertion sort is extended simple insertion sort.
             * The main idea of this sort is to put elements larger
             * than an element called pin to the end of array (the
             * proper area for such elements). It avoids expensive
             * movements of these elements through the whole array.
             */
            long pin = a[end];

            for (int i, p = high; ++low < end; ) {
                long ai = a[i = low];

                if (ai < a[i - 1]) { // Small element

                    /*
                     * Insert small element into sorted part.
                     */
                    a[i] = a[--i];

                    while (ai < a[--i]) {
                        a[i + 1] = a[i];
                    }
                    a[i + 1] = ai;

                } else if (p > i && ai > pin) { // Large element

                    /*
                     * Find element smaller than pin.
                     */
                    while (a[--p] > pin);

                    /*
                     * Swap it with large element.
                     */
                    if (p > i) {
                        ai = a[p];
                        a[p] = a[i];
                    }

                    /*
                     * Insert small element into sorted part.
                     */
                    while (ai < a[--i]) {
                        a[i + 1] = a[i];
                    }
                    a[i + 1] = ai;
                }
            }

            /*
             * Continue with pair insertion sort on remain part.
             */
            for (int i; low < high; ++low) {
                long a1 = a[i = low], a2 = a[++low];

                /*
                 * Insert two elements per iteration: at first, insert the
                 * larger element and then insert the smaller element, but
                 * from the position where the larger element was inserted.
                 */
                if (a1 > a2) {

                    while (a1 < a[--i]) {
                        a[i + 2] = a[i];
                    }
                    a[++i + 1] = a1;

                    while (a2 < a[--i]) {
                        a[i + 1] = a[i];
                    }
                    a[i + 1] = a2;

                } else if (a1 < a[i - 1]) {

                    while (a2 < a[--i]) {
                        a[i + 2] = a[i];
                    }
                    a[++i + 1] = a2;

                    while (a1 < a[--i]) {
                        a[i + 1] = a[i];
                    }
                    a[i + 1] = a1;
                }
            }
        }
    }

    /**
     * Sorts the specified range of the array using insertion sort.
     *
     * @param a the array to be sorted
     * @param low the index of the first element, inclusive, to be sorted
     * @param high the index of the last element, exclusive, to be sorted
     */
    private static void insertionSort(long[] a, int low, int high) {
        for (int i, k = low; ++k < high; ) {
            long ai = a[i = k];

            if (ai < a[i - 1]) {
                while (--i >= low && ai < a[i]) {
                    a[i + 1] = a[i];
                }
                a[i + 1] = ai;
            }
        }
    }

    /**
     * Sorts the specified range of the array using heap sort.
     *
     * @param a the array to be sorted
     * @param low the index of the first element, inclusive, to be sorted
     * @param high the index of the last element, exclusive, to be sorted
     */
    private static void heapSort(long[] a, int low, int high) {
        for (int k = (low + high) >>> 1; k > low; ) {
            pushDown(a, --k, a[k], low, high);
        }
        while (--high > low) {
            long max = a[low];
            pushDown(a, low, a[high], low, high);
            a[high] = max;
        }
    }

    /**
     * Pushes specified element down during heap sort.
     *
     * @param a the given array
     * @param p the start index
     * @param value the given element
     * @param low the index of the first element, inclusive, to be sorted
     * @param high the index of the last element, exclusive, to be sorted
     */
    private static void pushDown(long[] a, int p, long value, int low, int high) {
        for (int k ;; a[p] = a[p = k]) {
            k = (p << 1) - low + 2; // Index of the right child

            if (k > high) {
                break;
            }
            if (k == high || a[k] < a[k - 1]) {
                --k;
            }
            if (a[k] <= value) {
                break;
            }
        }
        a[p] = value;
    }

    /**
     * Tries to sort the specified range of the array.
     *
     * @param sorter parallel context
     * @param a the array to be sorted
     * @param low the index of the first element to be sorted
     * @param size the array size
     * @return true if finally sorted, false otherwise
     */
    private static boolean tryMergeRuns(Sorter sorter, long[] a, int low, int size) {

        /*
         * The run array is constructed only if initial runs are
         * long enough to continue, run[i] then holds start index
         * of the i-th sequence of elements in non-descending order.
         */
        int[] run = null;
        int high = low + size;
        int count = 1, last = low;

        /*
         * Identify all possible runs.
         */
        for (int k = low + 1; k < high; ) {

            /*
             * Find the end index of the current run.
             */
            if (a[k - 1] < a[k]) {

                // Identify ascending sequence
                while (++k < high && a[k - 1] <= a[k]);

            } else if (a[k - 1] > a[k]) {

                // Identify descending sequence
                while (++k < high && a[k - 1] >= a[k]);

                // Reverse into ascending order
                for (int i = last - 1, j = k; ++i < --j && a[i] > a[j]; ) {
                    long ai = a[i]; a[i] = a[j]; a[j] = ai;
                }
            } else { // Identify constant sequence
                for (long ak = a[k]; ++k < high && ak == a[k]; );

                if (k < high) {
                    continue;
                }
            }

            /*
             * Check special cases.
             */
            if (run == null) {
                if (k == high) {

                    /*
                     * The array is monotonous sequence,
                     * and therefore already sorted.
                     */
                    return true;
                }

                if (k - low < MIN_FIRST_RUN_SIZE) {

                    /*
                     * The first run is too small
                     * to proceed with scanning.
                     */
                    return false;
                }

                run = new int[((size >> 10) | 0x7F) & 0x3FF];
                run[0] = low;

            } else if (a[last - 1] > a[last]) {

                if (count > (k - low) >> MIN_FIRST_RUNS_FACTOR) {

                    /*
                     * The first runs are not long
                     * enough to continue scanning.
                     */
                    return false;
                }

                if (++count == MAX_RUN_CAPACITY) {

                    /*
                     * Array is not highly structured.
                     */
                    return false;
                }

                if (count == run.length) {

                    /*
                     * Increase capacity of index array.
                     */
                    run = Arrays.copyOf(run, count << 1);
                }
            }
            run[count] = (last = k);
        }

        /*
         * Merge runs of highly structured array.
         */
        if (count > 1) {
            long[] b; int offset = low;

            if (sorter == null || (b = (long[]) sorter.b) == null) {
                b = new long[size];
            } else {
                offset = sorter.offset;
            }
            mergeRuns(a, b, offset, 1, sorter != null, run, 0, count);
        }
        return true;
    }

    /**
     * Merges the specified runs.
     *
     * @param a the source array
     * @param b the temporary buffer used in merging
     * @param offset the start index in the source, inclusive
     * @param aim specifies merging: to source ( > 0), buffer ( < 0) or any ( == 0)
     * @param parallel indicates whether merging is performed in parallel
     * @param run the start indexes of the runs, inclusive
     * @param lo the start index of the first run, inclusive
     * @param hi the start index of the last run, inclusive
     * @return the destination where runs are merged
     */
    private static long[] mergeRuns(long[] a, long[] b, int offset,
            int aim, boolean parallel, int[] run, int lo, int hi) {

        if (hi - lo == 1) {
            if (aim >= 0) {
                return a;
            }
            for (int i = run[hi], j = i - offset, low = run[lo]; i > low;
                b[--j] = a[--i]
            );
            return b;
        }

        /*
         * Split into approximately equal parts.
         */
        int mi = lo, rmi = (run[lo] + run[hi]) >>> 1;
        while (run[++mi + 1] <= rmi);

        /*
         * Merge the left and right parts.
         */
        long[] a1, a2;

        if (parallel && hi - lo > MIN_RUN_COUNT) {
            RunMerger merger = new RunMerger(a, b, offset, 0, run, mi, hi).forkMe();
            a1 = mergeRuns(a, b, offset, -aim, true, run, lo, mi);
            a2 = (long[]) merger.getDestination();
        } else {
            a1 = mergeRuns(a, b, offset, -aim, false, run, lo, mi);
            a2 = mergeRuns(a, b, offset,    0, false, run, mi, hi);
        }

        long[] dst = a1 == a ? b : a;

        int k   = a1 == a ? run[lo] - offset : run[lo];
        int lo1 = a1 == b ? run[lo] - offset : run[lo];
        int hi1 = a1 == b ? run[mi] - offset : run[mi];
        int lo2 = a2 == b ? run[mi] - offset : run[mi];
        int hi2 = a2 == b ? run[hi] - offset : run[hi];

        if (parallel) {
            new Merger(null, dst, k, a1, lo1, hi1, a2, lo2, hi2).invoke();
        } else {
            mergeParts(null, dst, k, a1, lo1, hi1, a2, lo2, hi2);
        }
        return dst;
    }

    /**
     * Merges the sorted parts.
     *
     * @param merger parallel context
     * @param dst the destination where parts are merged
     * @param k the start index of the destination, inclusive
     * @param a1 the first part
     * @param lo1 the start index of the first part, inclusive
     * @param hi1 the end index of the first part, exclusive
     * @param a2 the second part
     * @param lo2 the start index of the second part, inclusive
     * @param hi2 the end index of the second part, exclusive
     */
    private static void mergeParts(Merger merger, long[] dst, int k,
            long[] a1, int lo1, int hi1, long[] a2, int lo2, int hi2) {

        if (merger != null && a1 == a2) {

            while (true) {

                /*
                 * The first part must be larger.
                 */
                if (hi1 - lo1 < hi2 - lo2) {
                    int lo = lo1; lo1 = lo2; lo2 = lo;
                    int hi = hi1; hi1 = hi2; hi2 = hi;
                }

                /*
                 * Small parts will be merged sequentially.
                 */
                if (hi1 - lo1 < MIN_PARALLEL_MERGE_PARTS_SIZE) {
                    break;
                }

                /*
                 * Find the median of the larger part.
                 */
                int mi1 = (lo1 + hi1) >>> 1;
                long key = a1[mi1];
                int mi2 = hi2;

                /*
                 * Partition the smaller part.
                 */
                for (int loo = lo2; loo < mi2; ) {
                    int t = (loo + mi2) >>> 1;

                    if (key > a2[t]) {
                        loo = t + 1;
                    } else {
                        mi2 = t;
                    }
                }

                int d = mi2 - lo2 + mi1 - lo1;

                /*
                 * Merge the right sub-parts in parallel.
                 */
                merger.forkMerger(dst, k + d, a1, mi1, hi1, a2, mi2, hi2);

                /*
                 * Process the sub-left parts.
                 */
                hi1 = mi1;
                hi2 = mi2;
            }
        }

        /*
         * Merge small parts sequentially.
         */
        while (lo1 < hi1 && lo2 < hi2) {
            dst[k++] = a1[lo1] < a2[lo2] ? a1[lo1++] : a2[lo2++];
        }
        if (dst != a1 || k < lo1) {
            while (lo1 < hi1) {
                dst[k++] = a1[lo1++];
            }
        }
        if (dst != a2 || k < lo2) {
            while (lo2 < hi2) {
                dst[k++] = a2[lo2++];
            }
        }
    }

// [byte]

    /**
     * Sorts the specified range of the array using
     * counting sort or insertion sort.
     *
     * @param a the array to be sorted
     * @param low the index of the first element, inclusive, to be sorted
     * @param high the index of the last element, exclusive, to be sorted
     */
    static void sort(byte[] a, int low, int high) {
        if (high - low > MIN_BYTE_COUNTING_SORT_SIZE) {
            countingSort(a, low, high);
        } else {
            insertionSort(a, low, high);
        }
    }

    /**
     * Sorts the specified range of the array using insertion sort.
     *
     * @param a the array to be sorted
     * @param low the index of the first element, inclusive, to be sorted
     * @param high the index of the last element, exclusive, to be sorted
     */
    private static void insertionSort(byte[] a, int low, int high) {
        for (int i, k = low; ++k < high; ) {
            byte ai = a[i = k];

            if (ai < a[i - 1]) {
                while (--i >= low && ai < a[i]) {
                    a[i + 1] = a[i];
                }
                a[i + 1] = ai;
            }
        }
    }

    /**
     * The number of distinct byte values.
     */
    private static final int NUM_BYTE_VALUES = 1 << 8;

    /**
     * Max index of byte counter.
     */
    private static final int MAX_BYTE_INDEX = Byte.MAX_VALUE + NUM_BYTE_VALUES + 1;

    /**
     * Sorts the specified range of the array using counting sort.
     *
     * @param a the array to be sorted
     * @param low the index of the first element, inclusive, to be sorted
     * @param high the index of the last element, exclusive, to be sorted
     */
    private static void countingSort(byte[] a, int low, int high) {
        int[] count = new int[NUM_BYTE_VALUES];

        /*
         * Compute a histogram with the number of each values.
         */
        for (int i = high; i > low; ++count[a[--i] & 0xFF]);

        /*
         * Place values on their final positions.
         */
        if (high - low > NUM_BYTE_VALUES) {
            for (int i = MAX_BYTE_INDEX; --i > Byte.MAX_VALUE; ) {
                int value = i & 0xFF;

                for (low = high - count[value]; high > low;
                    a[--high] = (byte) value
                );
            }
        } else {
            for (int i = MAX_BYTE_INDEX; high > low; ) {
                while (count[--i & 0xFF] == 0);

                int value = i & 0xFF;
                int c = count[value];

                do {
                    a[--high] = (byte) value;
                } while (--c > 0);
            }
        }
    }

// [char]

    /**
     * Sorts the specified range of the array using
     * counting sort or Dual-Pivot Quicksort.
     *
     * @param a the array to be sorted
     * @param low the index of the first element, inclusive, to be sorted
     * @param high the index of the last element, exclusive, to be sorted
     */
    static void sort(char[] a, int low, int high) {
        if (high - low > MIN_SHORT_OR_CHAR_COUNTING_SORT_SIZE) {
            countingSort(a, low, high);
        } else {
            sort(a, 0, low, high);
        }
    }

    /**
     * Sorts the specified array using the Dual-Pivot Quicksort and/or
     * other sorts in special-cases, possibly with parallel partitions.
     *
     * @param a the array to be sorted
     * @param bits the combination of recursion depth and bit flag, where
     *        the right bit "0" indicates that array is the leftmost part
     * @param low the index of the first element, inclusive, to be sorted
     * @param high the index of the last element, exclusive, to be sorted
     */
    static void sort(char[] a, int bits, int low, int high) {
        while (true) {
            int end = high - 1, size = high - low;

            /*
             * Invoke insertion sort on small leftmost part.
             */
            if (size < MAX_INSERTION_SORT_SIZE) {
                insertionSort(a, low, high);
                return;
            }

            /*
             * Switch to counting sort if execution
             * time is becoming quadratic.
             */
            if ((bits += DELTA) > MAX_RECURSION_DEPTH) {
                countingSort(a, low, high);
                return;
            }

            /*
             * Use an inexpensive approximation of the golden ratio
             * to select five sample elements and determine pivots.
             */
            int step = (size >> 3) * 3 + 3;

            /*
             * Five elements around (and including) the central element
             * will be used for pivot selection as described below. The
             * unequal choice of spacing these elements was empirically
             * determined to work well on a wide variety of inputs.
             */
            int e1 = low + step;
            int e5 = end - step;
            int e3 = (e1 + e5) >>> 1;
            int e2 = (e1 + e3) >>> 1;
            int e4 = (e3 + e5) >>> 1;
            char a3 = a[e3];

            /*
             * Sort these elements in place by the combination
             * of 4-element sorting network and insertion sort.
             *
             *    5 ------o-----------o------------
             *            |           |
             *    4 ------|-----o-----o-----o------
             *            |     |           |
             *    2 ------o-----|-----o-----o------
             *                  |     |
             *    1 ------------o-----o------------
             */
            if (a[e5] < a[e2]) { char t = a[e5]; a[e5] = a[e2]; a[e2] = t; }
            if (a[e4] < a[e1]) { char t = a[e4]; a[e4] = a[e1]; a[e1] = t; }
            if (a[e5] < a[e4]) { char t = a[e5]; a[e5] = a[e4]; a[e4] = t; }
            if (a[e2] < a[e1]) { char t = a[e2]; a[e2] = a[e1]; a[e1] = t; }
            if (a[e4] < a[e2]) { char t = a[e4]; a[e4] = a[e2]; a[e2] = t; }

            if (a3 < a[e2]) {
                if (a3 < a[e1]) {
                    a[e3] = a[e2]; a[e2] = a[e1]; a[e1] = a3;
                } else {
                    a[e3] = a[e2]; a[e2] = a3;
                }
            } else if (a3 > a[e4]) {
                if (a3 > a[e5]) {
                    a[e3] = a[e4]; a[e4] = a[e5]; a[e5] = a3;
                } else {
                    a[e3] = a[e4]; a[e4] = a3;
                }
            }

            // Pointers
            int lower = low; // The index of the last element of the left part
            int upper = end; // The index of the first element of the right part

            /*
             * Partitioning with 2 pivots in case of different elements.
             */
            if (a[e1] < a[e2] && a[e2] < a[e3] && a[e3] < a[e4] && a[e4] < a[e5]) {

                /*
                 * Use the first and fifth of the five sorted elements as
                 * the pivots. These values are inexpensive approximation
                 * of tertiles. Note, that pivot1 < pivot2.
                 */
                char pivot1 = a[e1];
                char pivot2 = a[e5];

                /*
                 * The first and the last elements to be sorted are moved
                 * to the locations formerly occupied by the pivots. When
                 * partitioning is completed, the pivots are swapped back
                 * into their final positions, and excluded from the next
                 * subsequent sorting.
                 */
                a[e1] = a[lower];
                a[e5] = a[upper];

                /*
                 * Skip elements, which are less or greater than the pivots.
                 */
                while (a[++lower] < pivot1);
                while (a[--upper] > pivot2);

                /*
                 * Backward 3-interval partitioning
                 *
                 *   left part                 central part          right part
                 * +------------------------------------------------------------+
                 * |  < pivot1  |   ?   |  pivot1 <= && <= pivot2  |  > pivot2  |
                 * +------------------------------------------------------------+
                 *             ^       ^                            ^
                 *             |       |                            |
                 *           lower     k                          upper
                 *
                 * Invariants:
                 *
                 *              all in (low, lower] < pivot1
                 *    pivot1 <= all in (k, upper)  <= pivot2
                 *              all in [upper, end) > pivot2
                 *
                 * Pointer k is the last index of ?-part
                 */
                for (int unused = --lower, k = ++upper; --k > lower; ) {
                    char ak = a[k];

                    if (ak < pivot1) { // Move a[k] to the left side
                        while (lower < k) {
                            if (a[++lower] >= pivot1) {
                                if (a[lower] > pivot2) {
                                    a[k] = a[--upper];
                                    a[upper] = a[lower];
                                } else {
                                    a[k] = a[lower];
                                }
                                a[lower] = ak;
                                break;
                            }
                        }
                    } else if (ak > pivot2) { // Move a[k] to the right side
                        a[k] = a[--upper];
                        a[upper] = ak;
                    }
                }

                /*
                 * Swap the pivots into their final positions.
                 */
                a[low] = a[lower]; a[lower] = pivot1;
                a[end] = a[upper]; a[upper] = pivot2;

                /*
                 * Sort non-left parts recursively,
                 * excluding known pivots.
                 */
                sort(a, bits | 1, lower + 1, upper);
                sort(a, bits | 1, upper + 1, high);

            } else { // Use single pivot in case of many equal elements

                /*
                 * Use the third of the five sorted elements as the pivot.
                 * This value is inexpensive approximation of the median.
                 */
                char pivot = a[e3];

                /*
                 * The first element to be sorted is moved to the
                 * location formerly occupied by the pivot. After
                 * completion of partitioning the pivot is swapped
                 * back into its final position, and excluded from
                 * the next subsequent sorting.
                 */
                a[e3] = a[lower];

                /*
                 * Traditional 3-way (Dutch National Flag) partitioning
                 *
                 *   left part                 central part    right part
                 * +------------------------------------------------------+
                 * |   < pivot   |     ?     |   == pivot   |   > pivot   |
                 * +------------------------------------------------------+
                 *              ^           ^                ^
                 *              |           |                |
                 *            lower         k              upper
                 *
                 * Invariants:
                 *
                 *   all in (low, lower] < pivot
                 *   all in (k, upper)  == pivot
                 *   all in [upper, end] > pivot
                 *
                 * Pointer k is the last index of ?-part
                 */
                for (int k = ++upper; --k > lower; ) {
                    char ak = a[k];

                    if (ak != pivot) {
                        a[k] = pivot;

                        if (ak < pivot) { // Move a[k] to the left side
                            while (a[++lower] < pivot);

                            if (a[lower] > pivot) {
                                a[--upper] = a[lower];
                            }
                            a[lower] = ak;
                        } else { // ak > pivot - Move a[k] to the right side
                            a[--upper] = ak;
                        }
                    }
                }

                /*
                 * Swap the pivot into its final position.
                 */
                a[low] = a[lower]; a[lower] = pivot;

                /*
                 * Sort the right part, excluding known pivot.
                 * All elements from the central part are
                 * equal and therefore already sorted.
                 */
                sort(a, bits | 1, upper, high);
            }
            high = lower; // Iterate along the left part
        }
    }

    /**
     * Sorts the specified range of the array using insertion sort.
     *
     * @param a the array to be sorted
     * @param low the index of the first element, inclusive, to be sorted
     * @param high the index of the last element, exclusive, to be sorted
     */
    private static void insertionSort(char[] a, int low, int high) {
        for (int i, k = low; ++k < high; ) {
            char ai = a[i = k];

            if (ai < a[i - 1]) {
                while (--i >= low && ai < a[i]) {
                    a[i + 1] = a[i];
                }
                a[i + 1] = ai;
            }
        }
    }

    /**
     * The number of distinct char values.
     */
    private static final int NUM_CHAR_VALUES = 1 << 16;

    /**
     * Sorts the specified range of the array using counting sort.
     *
     * @param a the array to be sorted
     * @param low the index of the first element, inclusive, to be sorted
     * @param high the index of the last element, exclusive, to be sorted
     */
    private static void countingSort(char[] a, int low, int high) {
        int[] count = new int[NUM_CHAR_VALUES];

        /*
         * Compute a histogram with the number of each values.
         */
        for (int i = high; i > low; ++count[a[--i]]);

        /*
         * Place values on their final positions.
         */
        if (high - low > NUM_CHAR_VALUES) {
            for (int i = NUM_CHAR_VALUES; i > 0; ) {
                for (low = high - count[--i]; high > low;
                    a[--high] = (char) i
                );
            }
        } else {
            for (int i = NUM_CHAR_VALUES; high > low; ) {
                while (count[--i] == 0);
                int c = count[i];

                do {
                    a[--high] = (char) i;
                } while (--c > 0);
            }
        }
    }

// [short]

    /**
     * Sorts the specified range of the array using
     * counting sort or Dual-Pivot Quicksort.
     *
     * @param a the array to be sorted
     * @param low the index of the first element, inclusive, to be sorted
     * @param high the index of the last element, exclusive, to be sorted
     */
    static void sort(short[] a, int low, int high) {
        if (high - low > MIN_SHORT_OR_CHAR_COUNTING_SORT_SIZE) {
            countingSort(a, low, high);
        } else {
            sort(a, 0, low, high);
        }
    }

    /**
     * Sorts the specified array using the Dual-Pivot Quicksort and/or
     * other sorts in special-cases, possibly with parallel partitions.
     *
     * @param a the array to be sorted
     * @param bits the combination of recursion depth and bit flag, where
     *        the right bit "0" indicates that array is the leftmost part
     * @param low the index of the first element, inclusive, to be sorted
     * @param high the index of the last element, exclusive, to be sorted
     */
    static void sort(short[] a, int bits, int low, int high) {
        while (true) {
            int end = high - 1, size = high - low;

            /*
             * Invoke insertion sort on small leftmost part.
             */
            if (size < MAX_INSERTION_SORT_SIZE) {
                insertionSort(a, low, high);
                return;
            }

            /*
             * Switch to counting sort if execution
             * time is becoming quadratic.
             */
            if ((bits += DELTA) > MAX_RECURSION_DEPTH) {
                countingSort(a, low, high);
                return;
            }

            /*
             * Use an inexpensive approximation of the golden ratio
             * to select five sample elements and determine pivots.
             */
            int step = (size >> 3) * 3 + 3;

            /*
             * Five elements around (and including) the central element
             * will be used for pivot selection as described below. The
             * unequal choice of spacing these elements was empirically
             * determined to work well on a wide variety of inputs.
             */
            int e1 = low + step;
            int e5 = end - step;
            int e3 = (e1 + e5) >>> 1;
            int e2 = (e1 + e3) >>> 1;
            int e4 = (e3 + e5) >>> 1;
            short a3 = a[e3];

            /*
             * Sort these elements in place by the combination
             * of 4-element sorting network and insertion sort.
             *
             *    5 ------o-----------o------------
             *            |           |
             *    4 ------|-----o-----o-----o------
             *            |     |           |
             *    2 ------o-----|-----o-----o------
             *                  |     |
             *    1 ------------o-----o------------
             */
            if (a[e5] < a[e2]) { short t = a[e5]; a[e5] = a[e2]; a[e2] = t; }
            if (a[e4] < a[e1]) { short t = a[e4]; a[e4] = a[e1]; a[e1] = t; }
            if (a[e5] < a[e4]) { short t = a[e5]; a[e5] = a[e4]; a[e4] = t; }
            if (a[e2] < a[e1]) { short t = a[e2]; a[e2] = a[e1]; a[e1] = t; }
            if (a[e4] < a[e2]) { short t = a[e4]; a[e4] = a[e2]; a[e2] = t; }

            if (a3 < a[e2]) {
                if (a3 < a[e1]) {
                    a[e3] = a[e2]; a[e2] = a[e1]; a[e1] = a3;
                } else {
                    a[e3] = a[e2]; a[e2] = a3;
                }
            } else if (a3 > a[e4]) {
                if (a3 > a[e5]) {
                    a[e3] = a[e4]; a[e4] = a[e5]; a[e5] = a3;
                } else {
                    a[e3] = a[e4]; a[e4] = a3;
                }
            }

            // Pointers
            int lower = low; // The index of the last element of the left part
            int upper = end; // The index of the first element of the right part

            /*
             * Partitioning with 2 pivots in case of different elements.
             */
            if (a[e1] < a[e2] && a[e2] < a[e3] && a[e3] < a[e4] && a[e4] < a[e5]) {

                /*
                 * Use the first and fifth of the five sorted elements as
                 * the pivots. These values are inexpensive approximation
                 * of tertiles. Note, that pivot1 < pivot2.
                 */
                short pivot1 = a[e1];
                short pivot2 = a[e5];

                /*
                 * The first and the last elements to be sorted are moved
                 * to the locations formerly occupied by the pivots. When
                 * partitioning is completed, the pivots are swapped back
                 * into their final positions, and excluded from the next
                 * subsequent sorting.
                 */
                a[e1] = a[lower];
                a[e5] = a[upper];

                /*
                 * Skip elements, which are less or greater than the pivots.
                 */
                while (a[++lower] < pivot1);
                while (a[--upper] > pivot2);

                /*
                 * Backward 3-interval partitioning
                 *
                 *   left part                 central part          right part
                 * +------------------------------------------------------------+
                 * |  < pivot1  |   ?   |  pivot1 <= && <= pivot2  |  > pivot2  |
                 * +------------------------------------------------------------+
                 *             ^       ^                            ^
                 *             |       |                            |
                 *           lower     k                          upper
                 *
                 * Invariants:
                 *
                 *              all in (low, lower] < pivot1
                 *    pivot1 <= all in (k, upper)  <= pivot2
                 *              all in [upper, end) > pivot2
                 *
                 * Pointer k is the last index of ?-part
                 */
                for (int unused = --lower, k = ++upper; --k > lower; ) {
                    short ak = a[k];

                    if (ak < pivot1) { // Move a[k] to the left side
                        while (lower < k) {
                            if (a[++lower] >= pivot1) {
                                if (a[lower] > pivot2) {
                                    a[k] = a[--upper];
                                    a[upper] = a[lower];
                                } else {
                                    a[k] = a[lower];
                                }
                                a[lower] = ak;
                                break;
                            }
                        }
                    } else if (ak > pivot2) { // Move a[k] to the right side
                        a[k] = a[--upper];
                        a[upper] = ak;
                    }
                }

                /*
                 * Swap the pivots into their final positions.
                 */
                a[low] = a[lower]; a[lower] = pivot1;
                a[end] = a[upper]; a[upper] = pivot2;

                /*
                 * Sort non-left parts recursively,
                 * excluding known pivots.
                 */
                sort(a, bits | 1, lower + 1, upper);
                sort(a, bits | 1, upper + 1, high);

            } else { // Use single pivot in case of many equal elements

                /*
                 * Use the third of the five sorted elements as the pivot.
                 * This value is inexpensive approximation of the median.
                 */
                short pivot = a[e3];

                /*
                 * The first element to be sorted is moved to the
                 * location formerly occupied by the pivot. After
                 * completion of partitioning the pivot is swapped
                 * back into its final position, and excluded from
                 * the next subsequent sorting.
                 */
                a[e3] = a[lower];

                /*
                 * Traditional 3-way (Dutch National Flag) partitioning
                 *
                 *   left part                 central part    right part
                 * +------------------------------------------------------+
                 * |   < pivot   |     ?     |   == pivot   |   > pivot   |
                 * +------------------------------------------------------+
                 *              ^           ^                ^
                 *              |           |                |
                 *            lower         k              upper
                 *
                 * Invariants:
                 *
                 *   all in (low, lower] < pivot
                 *   all in (k, upper)  == pivot
                 *   all in [upper, end] > pivot
                 *
                 * Pointer k is the last index of ?-part
                 */
                for (int k = ++upper; --k > lower; ) {
                    short ak = a[k];

                    if (ak != pivot) {
                        a[k] = pivot;

                        if (ak < pivot) { // Move a[k] to the left side
                            while (a[++lower] < pivot);

                            if (a[lower] > pivot) {
                                a[--upper] = a[lower];
                            }
                            a[lower] = ak;
                        } else { // ak > pivot - Move a[k] to the right side
                            a[--upper] = ak;
                        }
                    }
                }

                /*
                 * Swap the pivot into its final position.
                 */
                a[low] = a[lower]; a[lower] = pivot;

                /*
                 * Sort the right part, excluding known pivot.
                 * All elements from the central part are
                 * equal and therefore already sorted.
                 */
                sort(a, bits | 1, upper, high);
            }
            high = lower; // Iterate along the left part
        }
    }

    /**
     * Sorts the specified range of the array using insertion sort.
     *
     * @param a the array to be sorted
     * @param low the index of the first element, inclusive, to be sorted
     * @param high the index of the last element, exclusive, to be sorted
     */
    private static void insertionSort(short[] a, int low, int high) {
        for (int i, k = low; ++k < high; ) {
            short ai = a[i = k];

            if (ai < a[i - 1]) {
                while (--i >= low && ai < a[i]) {
                    a[i + 1] = a[i];
                }
                a[i + 1] = ai;
            }
        }
    }

    /**
     * The number of distinct short values.
     */
    private static final int NUM_SHORT_VALUES = 1 << 16;

    /**
     * Max index of short counter.
     */
    private static final int MAX_SHORT_INDEX = Short.MAX_VALUE + NUM_SHORT_VALUES + 1;

    /**
     * Sorts the specified range of the array using counting sort.
     *
     * @param a the array to be sorted
     * @param low the index of the first element, inclusive, to be sorted
     * @param high the index of the last element, exclusive, to be sorted
     */
    private static void countingSort(short[] a, int low, int high) {
        int[] count = new int[NUM_SHORT_VALUES];

        /*
         * Compute a histogram with the number of each values.
         */
        for (int i = high; i > low; ++count[a[--i] & 0xFFFF]);

        /*
         * Place values on their final positions.
         */
        if (high - low > NUM_SHORT_VALUES) {
            for (int i = MAX_SHORT_INDEX; --i > Short.MAX_VALUE; ) {
                int value = i & 0xFFFF;

                for (low = high - count[value]; high > low;
                    a[--high] = (short) value
                );
            }
        } else {
            for (int i = MAX_SHORT_INDEX; high > low; ) {
                while (count[--i & 0xFFFF] == 0);

                int value = i & 0xFFFF;
                int c = count[value];

                do {
                    a[--high] = (short) value;
                } while (--c > 0);
            }
        }
    }

// [float]

    /**
     * Sorts the specified range of the array using parallel merge
     * sort and/or Dual-Pivot Quicksort.
     *
     * To balance the faster splitting and parallelism of merge sort
     * with the faster element partitioning of Quicksort, ranges are
     * subdivided in tiers such that, if there is enough parallelism,
     * the four-way parallel merge is started, still ensuring enough
     * parallelism to process the partitions.
     *
     * @param a the array to be sorted
     * @param parallelism the parallelism level
     * @param low the index of the first element, inclusive, to be sorted
     * @param high the index of the last element, exclusive, to be sorted
     */
    static void sort(float[] a, int parallelism, int low, int high) {
        /*
         * Phase 1. Count the number of negative zero -0.0f,
         * turn them into positive zero, and move all NaNs
         * to the end of the array.
         */
        int numNegativeZero = 0;

        for (int k = high; k > low; ) {
            float ak = a[--k];

            if (ak == 0.0f && Float.floatToRawIntBits(ak) < 0) { // ak is -0.0f
                numNegativeZero += 1;
                a[k] = 0.0f;
            } else if (ak != ak) { // ak is NaN
                a[k] = a[--high];
                a[high] = ak;
            }
        }

        /*
         * Phase 2. Sort everything except NaNs,
         * which are already in place.
         */
        int size = high - low;

        if (parallelism > 1 && size > MIN_PARALLEL_SORT_SIZE) {
            int depth = getDepth(parallelism, size >> 12);
            float[] b = depth == 0 ? null : new float[size];
            new Sorter(null, a, b, low, size, low, depth).invoke();
        } else {
            sort(null, a, 0, low, high);
        }

        /*
         * Phase 3. Turn positive zero 0.0f
         * back into negative zero -0.0f.
         */
        if (++numNegativeZero == 1) {
            return;
        }

        /*
         * Find the position one less than
         * the index of the first zero.
         */
        while (low <= high) {
            int middle = (low + high) >>> 1;

            if (a[middle] < 0) {
                low = middle + 1;
            } else {
                high = middle - 1;
            }
        }

        /*
         * Replace the required number of 0.0f by -0.0f.
         */
        while (--numNegativeZero > 0) {
            a[++high] = -0.0f;
        }
    }

    /**
     * Sorts the specified array using the Dual-Pivot Quicksort and/or
     * other sorts in special-cases, possibly with parallel partitions.
     *
     * @param sorter parallel context
     * @param a the array to be sorted
     * @param bits the combination of recursion depth and bit flag, where
     *        the right bit "0" indicates that array is the leftmost part
     * @param low the index of the first element, inclusive, to be sorted
     * @param high the index of the last element, exclusive, to be sorted
     */
    static void sort(Sorter sorter, float[] a, int bits, int low, int high) {
        while (true) {
            int end = high - 1, size = high - low;

            /*
             * Run mixed insertion sort on small non-leftmost parts.
             */
            if (size < MAX_MIXED_INSERTION_SORT_SIZE + bits && (bits & 1) > 0) {
                mixedInsertionSort(a, low, high - 3 * ((size >> 5) << 3), high);
                return;
            }

            /*
             * Invoke insertion sort on small leftmost part.
             */
            if (size < MAX_INSERTION_SORT_SIZE) {
                insertionSort(a, low, high);
                return;
            }

            /*
             * Check if the whole array or large non-leftmost
             * parts are nearly sorted and then merge runs.
             */
            if ((bits == 0 || size > MIN_TRY_MERGE_SIZE && (bits & 1) > 0)
                    && tryMergeRuns(sorter, a, low, size)) {
                return;
            }

            /*
             * Switch to heap sort if execution
             * time is becoming quadratic.
             */
            if ((bits += DELTA) > MAX_RECURSION_DEPTH) {
                heapSort(a, low, high);
                return;
            }

            /*
             * Use an inexpensive approximation of the golden ratio
             * to select five sample elements and determine pivots.
             */
            int step = (size >> 3) * 3 + 3;

            /*
             * Five elements around (and including) the central element
             * will be used for pivot selection as described below. The
             * unequal choice of spacing these elements was empirically
             * determined to work well on a wide variety of inputs.
             */
            int e1 = low + step;
            int e5 = end - step;
            int e3 = (e1 + e5) >>> 1;
            int e2 = (e1 + e3) >>> 1;
            int e4 = (e3 + e5) >>> 1;
            float a3 = a[e3];

            /*
             * Sort these elements in place by the combination
             * of 4-element sorting network and insertion sort.
             *
             *    5 ------o-----------o------------
             *            |           |
             *    4 ------|-----o-----o-----o------
             *            |     |           |
             *    2 ------o-----|-----o-----o------
             *                  |     |
             *    1 ------------o-----o------------
             */
            if (a[e5] < a[e2]) { float t = a[e5]; a[e5] = a[e2]; a[e2] = t; }
            if (a[e4] < a[e1]) { float t = a[e4]; a[e4] = a[e1]; a[e1] = t; }
            if (a[e5] < a[e4]) { float t = a[e5]; a[e5] = a[e4]; a[e4] = t; }
            if (a[e2] < a[e1]) { float t = a[e2]; a[e2] = a[e1]; a[e1] = t; }
            if (a[e4] < a[e2]) { float t = a[e4]; a[e4] = a[e2]; a[e2] = t; }

            if (a3 < a[e2]) {
                if (a3 < a[e1]) {
                    a[e3] = a[e2]; a[e2] = a[e1]; a[e1] = a3;
                } else {
                    a[e3] = a[e2]; a[e2] = a3;
                }
            } else if (a3 > a[e4]) {
                if (a3 > a[e5]) {
                    a[e3] = a[e4]; a[e4] = a[e5]; a[e5] = a3;
                } else {
                    a[e3] = a[e4]; a[e4] = a3;
                }
            }

            // Pointers
            int lower = low; // The index of the last element of the left part
            int upper = end; // The index of the first element of the right part

            /*
             * Partitioning with 2 pivots in case of different elements.
             */
            if (a[e1] < a[e2] && a[e2] < a[e3] && a[e3] < a[e4] && a[e4] < a[e5]) {

                /*
                 * Use the first and fifth of the five sorted elements as
                 * the pivots. These values are inexpensive approximation
                 * of tertiles. Note, that pivot1 < pivot2.
                 */
                float pivot1 = a[e1];
                float pivot2 = a[e5];

                /*
                 * The first and the last elements to be sorted are moved
                 * to the locations formerly occupied by the pivots. When
                 * partitioning is completed, the pivots are swapped back
                 * into their final positions, and excluded from the next
                 * subsequent sorting.
                 */
                a[e1] = a[lower];
                a[e5] = a[upper];

                /*
                 * Skip elements, which are less or greater than the pivots.
                 */
                while (a[++lower] < pivot1);
                while (a[--upper] > pivot2);

                /*
                 * Backward 3-interval partitioning
                 *
                 *   left part                 central part          right part
                 * +------------------------------------------------------------+
                 * |  < pivot1  |   ?   |  pivot1 <= && <= pivot2  |  > pivot2  |
                 * +------------------------------------------------------------+
                 *             ^       ^                            ^
                 *             |       |                            |
                 *           lower     k                          upper
                 *
                 * Invariants:
                 *
                 *              all in (low, lower] < pivot1
                 *    pivot1 <= all in (k, upper)  <= pivot2
                 *              all in [upper, end) > pivot2
                 *
                 * Pointer k is the last index of ?-part
                 */
                for (int unused = --lower, k = ++upper; --k > lower; ) {
                    float ak = a[k];

                    if (ak < pivot1) { // Move a[k] to the left side
                        while (lower < k) {
                            if (a[++lower] >= pivot1) {
                                if (a[lower] > pivot2) {
                                    a[k] = a[--upper];
                                    a[upper] = a[lower];
                                } else {
                                    a[k] = a[lower];
                                }
                                a[lower] = ak;
                                break;
                            }
                        }
                    } else if (ak > pivot2) { // Move a[k] to the right side
                        a[k] = a[--upper];
                        a[upper] = ak;
                    }
                }

                /*
                 * Swap the pivots into their final positions.
                 */
                a[low] = a[lower]; a[lower] = pivot1;
                a[end] = a[upper]; a[upper] = pivot2;

                /*
                 * Sort non-left parts recursively (possibly in parallel),
                 * excluding known pivots.
                 */
                if (size > MIN_PARALLEL_SORT_SIZE && sorter != null) {
                    sorter.forkSorter(bits | 1, lower + 1, upper);
                    sorter.forkSorter(bits | 1, upper + 1, high);
                } else {
                    sort(sorter, a, bits | 1, lower + 1, upper);
                    sort(sorter, a, bits | 1, upper + 1, high);
                }

            } else { // Use single pivot in case of many equal elements

                /*
                 * Use the third of the five sorted elements as the pivot.
                 * This value is inexpensive approximation of the median.
                 */
                float pivot = a[e3];

                /*
                 * The first element to be sorted is moved to the
                 * location formerly occupied by the pivot. After
                 * completion of partitioning the pivot is swapped
                 * back into its final position, and excluded from
                 * the next subsequent sorting.
                 */
                a[e3] = a[lower];

                /*
                 * Traditional 3-way (Dutch National Flag) partitioning
                 *
                 *   left part                 central part    right part
                 * +------------------------------------------------------+
                 * |   < pivot   |     ?     |   == pivot   |   > pivot   |
                 * +------------------------------------------------------+
                 *              ^           ^                ^
                 *              |           |                |
                 *            lower         k              upper
                 *
                 * Invariants:
                 *
                 *   all in (low, lower] < pivot
                 *   all in (k, upper)  == pivot
                 *   all in [upper, end] > pivot
                 *
                 * Pointer k is the last index of ?-part
                 */
                for (int k = ++upper; --k > lower; ) {
                    float ak = a[k];

                    if (ak != pivot) {
                        a[k] = pivot;

                        if (ak < pivot) { // Move a[k] to the left side
                            while (a[++lower] < pivot);

                            if (a[lower] > pivot) {
                                a[--upper] = a[lower];
                            }
                            a[lower] = ak;
                        } else { // ak > pivot - Move a[k] to the right side
                            a[--upper] = ak;
                        }
                    }
                }

                /*
                 * Swap the pivot into its final position.
                 */
                a[low] = a[lower]; a[lower] = pivot;

                /*
                 * Sort the right part (possibly in parallel), excluding
                 * known pivot. All elements from the central part are
                 * equal and therefore already sorted.
                 */
                if (size > MIN_PARALLEL_SORT_SIZE && sorter != null) {
                    sorter.forkSorter(bits | 1, upper, high);
                } else {
                    sort(sorter, a, bits | 1, upper, high);
                }
            }
            high = lower; // Iterate along the left part
        }
    }

    /**
     * Sorts the specified range of the array using mixed insertion sort.
     *
     * Mixed insertion sort is combination of simple insertion sort,
     * pin insertion sort and pair insertion sort.
     *
     * In the context of Dual-Pivot Quicksort, the pivot element
     * from the left part plays the role of sentinel, because it
     * is less than any elements from the given part. Therefore,
     * expensive check of the left range can be skipped on each
     * iteration unless it is the leftmost call.
     *
     * @param a the array to be sorted
     * @param low the index of the first element, inclusive, to be sorted
     * @param end the index of the last element for simple insertion sort
     * @param high the index of the last element, exclusive, to be sorted
     */
    private static void mixedInsertionSort(float[] a, int low, int end, int high) {
        if (end == high) {

            /*
             * Invoke simple insertion sort on tiny array.
             */
            for (int i; ++low < end; ) {
                float ai = a[i = low];

                while (ai < a[--i]) {
                    a[i + 1] = a[i];
                }
                a[i + 1] = ai;
            }
        } else {

            /*
             * Start with pin insertion sort on small part.
             *
             * Pin insertion sort is extended simple insertion sort.
             * The main idea of this sort is to put elements larger
             * than an element called pin to the end of array (the
             * proper area for such elements). It avoids expensive
             * movements of these elements through the whole array.
             */
            float pin = a[end];

            for (int i, p = high; ++low < end; ) {
                float ai = a[i = low];

                if (ai < a[i - 1]) { // Small element

                    /*
                     * Insert small element into sorted part.
                     */
                    a[i] = a[--i];

                    while (ai < a[--i]) {
                        a[i + 1] = a[i];
                    }
                    a[i + 1] = ai;

                } else if (p > i && ai > pin) { // Large element

                    /*
                     * Find element smaller than pin.
                     */
                    while (a[--p] > pin);

                    /*
                     * Swap it with large element.
                     */
                    if (p > i) {
                        ai = a[p];
                        a[p] = a[i];
                    }

                    /*
                     * Insert small element into sorted part.
                     */
                    while (ai < a[--i]) {
                        a[i + 1] = a[i];
                    }
                    a[i + 1] = ai;
                }
            }

            /*
             * Continue with pair insertion sort on remain part.
             */
            for (int i; low < high; ++low) {
                float a1 = a[i = low], a2 = a[++low];

                /*
                 * Insert two elements per iteration: at first, insert the
                 * larger element and then insert the smaller element, but
                 * from the position where the larger element was inserted.
                 */
                if (a1 > a2) {

                    while (a1 < a[--i]) {
                        a[i + 2] = a[i];
                    }
                    a[++i + 1] = a1;

                    while (a2 < a[--i]) {
                        a[i + 1] = a[i];
                    }
                    a[i + 1] = a2;

                } else if (a1 < a[i - 1]) {

                    while (a2 < a[--i]) {
                        a[i + 2] = a[i];
                    }
                    a[++i + 1] = a2;

                    while (a1 < a[--i]) {
                        a[i + 1] = a[i];
                    }
                    a[i + 1] = a1;
                }
            }
        }
    }

    /**
     * Sorts the specified range of the array using insertion sort.
     *
     * @param a the array to be sorted
     * @param low the index of the first element, inclusive, to be sorted
     * @param high the index of the last element, exclusive, to be sorted
     */
    private static void insertionSort(float[] a, int low, int high) {
        for (int i, k = low; ++k < high; ) {
            float ai = a[i = k];

            if (ai < a[i - 1]) {
                while (--i >= low && ai < a[i]) {
                    a[i + 1] = a[i];
                }
                a[i + 1] = ai;
            }
        }
    }

    /**
     * Sorts the specified range of the array using heap sort.
     *
     * @param a the array to be sorted
     * @param low the index of the first element, inclusive, to be sorted
     * @param high the index of the last element, exclusive, to be sorted
     */
    private static void heapSort(float[] a, int low, int high) {
        for (int k = (low + high) >>> 1; k > low; ) {
            pushDown(a, --k, a[k], low, high);
        }
        while (--high > low) {
            float max = a[low];
            pushDown(a, low, a[high], low, high);
            a[high] = max;
        }
    }

    /**
     * Pushes specified element down during heap sort.
     *
     * @param a the given array
     * @param p the start index
     * @param value the given element
     * @param low the index of the first element, inclusive, to be sorted
     * @param high the index of the last element, exclusive, to be sorted
     */
    private static void pushDown(float[] a, int p, float value, int low, int high) {
        for (int k ;; a[p] = a[p = k]) {
            k = (p << 1) - low + 2; // Index of the right child

            if (k > high) {
                break;
            }
            if (k == high || a[k] < a[k - 1]) {
                --k;
            }
            if (a[k] <= value) {
                break;
            }
        }
        a[p] = value;
    }

    /**
     * Tries to sort the specified range of the array.
     *
     * @param sorter parallel context
     * @param a the array to be sorted
     * @param low the index of the first element to be sorted
     * @param size the array size
     * @return true if finally sorted, false otherwise
     */
    private static boolean tryMergeRuns(Sorter sorter, float[] a, int low, int size) {

        /*
         * The run array is constructed only if initial runs are
         * long enough to continue, run[i] then holds start index
         * of the i-th sequence of elements in non-descending order.
         */
        int[] run = null;
        int high = low + size;
        int count = 1, last = low;

        /*
         * Identify all possible runs.
         */
        for (int k = low + 1; k < high; ) {

            /*
             * Find the end index of the current run.
             */
            if (a[k - 1] < a[k]) {

                // Identify ascending sequence
                while (++k < high && a[k - 1] <= a[k]);

            } else if (a[k - 1] > a[k]) {

                // Identify descending sequence
                while (++k < high && a[k - 1] >= a[k]);

                // Reverse into ascending order
                for (int i = last - 1, j = k; ++i < --j && a[i] > a[j]; ) {
                    float ai = a[i]; a[i] = a[j]; a[j] = ai;
                }
            } else { // Identify constant sequence
                for (float ak = a[k]; ++k < high && ak == a[k]; );

                if (k < high) {
                    continue;
                }
            }

            /*
             * Check special cases.
             */
            if (run == null) {
                if (k == high) {

                    /*
                     * The array is monotonous sequence,
                     * and therefore already sorted.
                     */
                    return true;
                }

                if (k - low < MIN_FIRST_RUN_SIZE) {

                    /*
                     * The first run is too small
                     * to proceed with scanning.
                     */
                    return false;
                }

                run = new int[((size >> 10) | 0x7F) & 0x3FF];
                run[0] = low;

            } else if (a[last - 1] > a[last]) {

                if (count > (k - low) >> MIN_FIRST_RUNS_FACTOR) {

                    /*
                     * The first runs are not long
                     * enough to continue scanning.
                     */
                    return false;
                }

                if (++count == MAX_RUN_CAPACITY) {

                    /*
                     * Array is not highly structured.
                     */
                    return false;
                }

                if (count == run.length) {

                    /*
                     * Increase capacity of index array.
                     */
                    run = Arrays.copyOf(run, count << 1);
                }
            }
            run[count] = (last = k);
        }

        /*
         * Merge runs of highly structured array.
         */
        if (count > 1) {
            float[] b; int offset = low;

            if (sorter == null || (b = (float[]) sorter.b) == null) {
                b = new float[size];
            } else {
                offset = sorter.offset;
            }
            mergeRuns(a, b, offset, 1, sorter != null, run, 0, count);
        }
        return true;
    }

    /**
     * Merges the specified runs.
     *
     * @param a the source array
     * @param b the temporary buffer used in merging
     * @param offset the start index in the source, inclusive
     * @param aim specifies merging: to source ( > 0), buffer ( < 0) or any ( == 0)
     * @param parallel indicates whether merging is performed in parallel
     * @param run the start indexes of the runs, inclusive
     * @param lo the start index of the first run, inclusive
     * @param hi the start index of the last run, inclusive
     * @return the destination where runs are merged
     */
    private static float[] mergeRuns(float[] a, float[] b, int offset,
            int aim, boolean parallel, int[] run, int lo, int hi) {

        if (hi - lo == 1) {
            if (aim >= 0) {
                return a;
            }
            for (int i = run[hi], j = i - offset, low = run[lo]; i > low;
                b[--j] = a[--i]
            );
            return b;
        }

        /*
         * Split into approximately equal parts.
         */
        int mi = lo, rmi = (run[lo] + run[hi]) >>> 1;
        while (run[++mi + 1] <= rmi);

        /*
         * Merge the left and right parts.
         */
        float[] a1, a2;

        if (parallel && hi - lo > MIN_RUN_COUNT) {
            RunMerger merger = new RunMerger(a, b, offset, 0, run, mi, hi).forkMe();
            a1 = mergeRuns(a, b, offset, -aim, true, run, lo, mi);
            a2 = (float[]) merger.getDestination();
        } else {
            a1 = mergeRuns(a, b, offset, -aim, false, run, lo, mi);
            a2 = mergeRuns(a, b, offset,    0, false, run, mi, hi);
        }

        float[] dst = a1 == a ? b : a;

        int k   = a1 == a ? run[lo] - offset : run[lo];
        int lo1 = a1 == b ? run[lo] - offset : run[lo];
        int hi1 = a1 == b ? run[mi] - offset : run[mi];
        int lo2 = a2 == b ? run[mi] - offset : run[mi];
        int hi2 = a2 == b ? run[hi] - offset : run[hi];

        if (parallel) {
            new Merger(null, dst, k, a1, lo1, hi1, a2, lo2, hi2).invoke();
        } else {
            mergeParts(null, dst, k, a1, lo1, hi1, a2, lo2, hi2);
        }
        return dst;
    }

    /**
     * Merges the sorted parts.
     *
     * @param merger parallel context
     * @param dst the destination where parts are merged
     * @param k the start index of the destination, inclusive
     * @param a1 the first part
     * @param lo1 the start index of the first part, inclusive
     * @param hi1 the end index of the first part, exclusive
     * @param a2 the second part
     * @param lo2 the start index of the second part, inclusive
     * @param hi2 the end index of the second part, exclusive
     */
    private static void mergeParts(Merger merger, float[] dst, int k,
            float[] a1, int lo1, int hi1, float[] a2, int lo2, int hi2) {

        if (merger != null && a1 == a2) {

            while (true) {

                /*
                 * The first part must be larger.
                 */
                if (hi1 - lo1 < hi2 - lo2) {
                    int lo = lo1; lo1 = lo2; lo2 = lo;
                    int hi = hi1; hi1 = hi2; hi2 = hi;
                }

                /*
                 * Small parts will be merged sequentially.
                 */
                if (hi1 - lo1 < MIN_PARALLEL_MERGE_PARTS_SIZE) {
                    break;
                }

                /*
                 * Find the median of the larger part.
                 */
                int mi1 = (lo1 + hi1) >>> 1;
                float key = a1[mi1];
                int mi2 = hi2;

                /*
                 * Partition the smaller part.
                 */
                for (int loo = lo2; loo < mi2; ) {
                    int t = (loo + mi2) >>> 1;

                    if (key > a2[t]) {
                        loo = t + 1;
                    } else {
                        mi2 = t;
                    }
                }

                int d = mi2 - lo2 + mi1 - lo1;

                /*
                 * Merge the right sub-parts in parallel.
                 */
                merger.forkMerger(dst, k + d, a1, mi1, hi1, a2, mi2, hi2);

                /*
                 * Process the sub-left parts.
                 */
                hi1 = mi1;
                hi2 = mi2;
            }
        }

        /*
         * Merge small parts sequentially.
         */
        while (lo1 < hi1 && lo2 < hi2) {
            dst[k++] = a1[lo1] < a2[lo2] ? a1[lo1++] : a2[lo2++];
        }
        if (dst != a1 || k < lo1) {
            while (lo1 < hi1) {
                dst[k++] = a1[lo1++];
            }
        }
        if (dst != a2 || k < lo2) {
            while (lo2 < hi2) {
                dst[k++] = a2[lo2++];
            }
        }
    }

// [double]

    /**
     * Sorts the specified range of the array using parallel merge
     * sort and/or Dual-Pivot Quicksort.
     *
     * To balance the faster splitting and parallelism of merge sort
     * with the faster element partitioning of Quicksort, ranges are
     * subdivided in tiers such that, if there is enough parallelism,
     * the four-way parallel merge is started, still ensuring enough
     * parallelism to process the partitions.
     *
     * @param a the array to be sorted
     * @param parallelism the parallelism level
     * @param low the index of the first element, inclusive, to be sorted
     * @param high the index of the last element, exclusive, to be sorted
     */
    static void sort(double[] a, int parallelism, int low, int high) {
        /*
         * Phase 1. Count the number of negative zero -0.0d,
         * turn them into positive zero, and move all NaNs
         * to the end of the array.
         */
        int numNegativeZero = 0;

        for (int k = high; k > low; ) {
            double ak = a[--k];

            if (ak == 0.0d && Double.doubleToRawLongBits(ak) < 0) { // ak is -0.0d
                numNegativeZero += 1;
                a[k] = 0.0d;
            } else if (ak != ak) { // ak is NaN
                a[k] = a[--high];
                a[high] = ak;
            }
        }

        /*
         * Phase 2. Sort everything except NaNs,
         * which are already in place.
         */
        int size = high - low;

        if (parallelism > 1 && size > MIN_PARALLEL_SORT_SIZE) {
            int depth = getDepth(parallelism, size >> 12);
            double[] b = depth == 0 ? null : new double[size];
            new Sorter(null, a, b, low, size, low, depth).invoke();
        } else {
            sort(null, a, 0, low, high);
        }

        /*
         * Phase 3. Turn positive zero 0.0d
         * back into negative zero -0.0d.
         */
        if (++numNegativeZero == 1) {
            return;
        }

        /*
         * Find the position one less than
         * the index of the first zero.
         */
        while (low <= high) {
            int middle = (low + high) >>> 1;

            if (a[middle] < 0) {
                low = middle + 1;
            } else {
                high = middle - 1;
            }
        }

        /*
         * Replace the required number of 0.0d by -0.0d.
         */
        while (--numNegativeZero > 0) {
            a[++high] = -0.0d;
        }
    }

    /**
     * Sorts the specified array using the Dual-Pivot Quicksort and/or
     * other sorts in special-cases, possibly with parallel partitions.
     *
     * @param sorter parallel context
     * @param a the array to be sorted
     * @param bits the combination of recursion depth and bit flag, where
     *        the right bit "0" indicates that array is the leftmost part
     * @param low the index of the first element, inclusive, to be sorted
     * @param high the index of the last element, exclusive, to be sorted
     */
    static void sort(Sorter sorter, double[] a, int bits, int low, int high) {
        while (true) {
            int end = high - 1, size = high - low;

            /*
             * Run mixed insertion sort on small non-leftmost parts.
             */
            if (size < MAX_MIXED_INSERTION_SORT_SIZE + bits && (bits & 1) > 0) {
                mixedInsertionSort(a, low, high - 3 * ((size >> 5) << 3), high);
                return;
            }

            /*
             * Invoke insertion sort on small leftmost part.
             */
            if (size < MAX_INSERTION_SORT_SIZE) {
                insertionSort(a, low, high);
                return;
            }

            /*
             * Check if the whole array or large non-leftmost
             * parts are nearly sorted and then merge runs.
             */
            if ((bits == 0 || size > MIN_TRY_MERGE_SIZE && (bits & 1) > 0)
                    && tryMergeRuns(sorter, a, low, size)) {
                return;
            }

            /*
             * Switch to heap sort if execution
             * time is becoming quadratic.
             */
            if ((bits += DELTA) > MAX_RECURSION_DEPTH) {
                heapSort(a, low, high);
                return;
            }

            /*
             * Use an inexpensive approximation of the golden ratio
             * to select five sample elements and determine pivots.
             */
            int step = (size >> 3) * 3 + 3;

            /*
             * Five elements around (and including) the central element
             * will be used for pivot selection as described below. The
             * unequal choice of spacing these elements was empirically
             * determined to work well on a wide variety of inputs.
             */
            int e1 = low + step;
            int e5 = end - step;
            int e3 = (e1 + e5) >>> 1;
            int e2 = (e1 + e3) >>> 1;
            int e4 = (e3 + e5) >>> 1;
            double a3 = a[e3];

            /*
             * Sort these elements in place by the combination
             * of 4-element sorting network and insertion sort.
             *
             *    5 ------o-----------o------------
             *            |           |
             *    4 ------|-----o-----o-----o------
             *            |     |           |
             *    2 ------o-----|-----o-----o------
             *                  |     |
             *    1 ------------o-----o------------
             */
            if (a[e5] < a[e2]) { double t = a[e5]; a[e5] = a[e2]; a[e2] = t; }
            if (a[e4] < a[e1]) { double t = a[e4]; a[e4] = a[e1]; a[e1] = t; }
            if (a[e5] < a[e4]) { double t = a[e5]; a[e5] = a[e4]; a[e4] = t; }
            if (a[e2] < a[e1]) { double t = a[e2]; a[e2] = a[e1]; a[e1] = t; }
            if (a[e4] < a[e2]) { double t = a[e4]; a[e4] = a[e2]; a[e2] = t; }

            if (a3 < a[e2]) {
                if (a3 < a[e1]) {
                    a[e3] = a[e2]; a[e2] = a[e1]; a[e1] = a3;
                } else {
                    a[e3] = a[e2]; a[e2] = a3;
                }
            } else if (a3 > a[e4]) {
                if (a3 > a[e5]) {
                    a[e3] = a[e4]; a[e4] = a[e5]; a[e5] = a3;
                } else {
                    a[e3] = a[e4]; a[e4] = a3;
                }
            }

            // Pointers
            int lower = low; // The index of the last element of the left part
            int upper = end; // The index of the first element of the right part

            /*
             * Partitioning with 2 pivots in case of different elements.
             */
            if (a[e1] < a[e2] && a[e2] < a[e3] && a[e3] < a[e4] && a[e4] < a[e5]) {

                /*
                 * Use the first and fifth of the five sorted elements as
                 * the pivots. These values are inexpensive approximation
                 * of tertiles. Note, that pivot1 < pivot2.
                 */
                double pivot1 = a[e1];
                double pivot2 = a[e5];

                /*
                 * The first and the last elements to be sorted are moved
                 * to the locations formerly occupied by the pivots. When
                 * partitioning is completed, the pivots are swapped back
                 * into their final positions, and excluded from the next
                 * subsequent sorting.
                 */
                a[e1] = a[lower];
                a[e5] = a[upper];

                /*
                 * Skip elements, which are less or greater than the pivots.
                 */
                while (a[++lower] < pivot1);
                while (a[--upper] > pivot2);

                /*
                 * Backward 3-interval partitioning
                 *
                 *   left part                 central part          right part
                 * +------------------------------------------------------------+
                 * |  < pivot1  |   ?   |  pivot1 <= && <= pivot2  |  > pivot2  |
                 * +------------------------------------------------------------+
                 *             ^       ^                            ^
                 *             |       |                            |
                 *           lower     k                          upper
                 *
                 * Invariants:
                 *
                 *              all in (low, lower] < pivot1
                 *    pivot1 <= all in (k, upper)  <= pivot2
                 *              all in [upper, end) > pivot2
                 *
                 * Pointer k is the last index of ?-part
                 */
                for (int unused = --lower, k = ++upper; --k > lower; ) {
                    double ak = a[k];

                    if (ak < pivot1) { // Move a[k] to the left side
                        while (lower < k) {
                            if (a[++lower] >= pivot1) {
                                if (a[lower] > pivot2) {
                                    a[k] = a[--upper];
                                    a[upper] = a[lower];
                                } else {
                                    a[k] = a[lower];
                                }
                                a[lower] = ak;
                                break;
                            }
                        }
                    } else if (ak > pivot2) { // Move a[k] to the right side
                        a[k] = a[--upper];
                        a[upper] = ak;
                    }
                }

                /*
                 * Swap the pivots into their final positions.
                 */
                a[low] = a[lower]; a[lower] = pivot1;
                a[end] = a[upper]; a[upper] = pivot2;

                /*
                 * Sort non-left parts recursively (possibly in parallel),
                 * excluding known pivots.
                 */
                if (size > MIN_PARALLEL_SORT_SIZE && sorter != null) {
                    sorter.forkSorter(bits | 1, lower + 1, upper);
                    sorter.forkSorter(bits | 1, upper + 1, high);
                } else {
                    sort(sorter, a, bits | 1, lower + 1, upper);
                    sort(sorter, a, bits | 1, upper + 1, high);
                }

            } else { // Use single pivot in case of many equal elements

                /*
                 * Use the third of the five sorted elements as the pivot.
                 * This value is inexpensive approximation of the median.
                 */
                double pivot = a[e3];

                /*
                 * The first element to be sorted is moved to the
                 * location formerly occupied by the pivot. After
                 * completion of partitioning the pivot is swapped
                 * back into its final position, and excluded from
                 * the next subsequent sorting.
                 */
                a[e3] = a[lower];

                /*
                 * Traditional 3-way (Dutch National Flag) partitioning
                 *
                 *   left part                 central part    right part
                 * +------------------------------------------------------+
                 * |   < pivot   |     ?     |   == pivot   |   > pivot   |
                 * +------------------------------------------------------+
                 *              ^           ^                ^
                 *              |           |                |
                 *            lower         k              upper
                 *
                 * Invariants:
                 *
                 *   all in (low, lower] < pivot
                 *   all in (k, upper)  == pivot
                 *   all in [upper, end] > pivot
                 *
                 * Pointer k is the last index of ?-part
                 */
                for (int k = ++upper; --k > lower; ) {
                    double ak = a[k];

                    if (ak != pivot) {
                        a[k] = pivot;

                        if (ak < pivot) { // Move a[k] to the left side
                            while (a[++lower] < pivot);

                            if (a[lower] > pivot) {
                                a[--upper] = a[lower];
                            }
                            a[lower] = ak;
                        } else { // ak > pivot - Move a[k] to the right side
                            a[--upper] = ak;
                        }
                    }
                }

                /*
                 * Swap the pivot into its final position.
                 */
                a[low] = a[lower]; a[lower] = pivot;

                /*
                 * Sort the right part (possibly in parallel), excluding
                 * known pivot. All elements from the central part are
                 * equal and therefore already sorted.
                 */
                if (size > MIN_PARALLEL_SORT_SIZE && sorter != null) {
                    sorter.forkSorter(bits | 1, upper, high);
                } else {
                    sort(sorter, a, bits | 1, upper, high);
                }
            }
            high = lower; // Iterate along the left part
        }
    }

    /**
     * Sorts the specified range of the array using mixed insertion sort.
     *
     * Mixed insertion sort is combination of simple insertion sort,
     * pin insertion sort and pair insertion sort.
     *
     * In the context of Dual-Pivot Quicksort, the pivot element
     * from the left part plays the role of sentinel, because it
     * is less than any elements from the given part. Therefore,
     * expensive check of the left range can be skipped on each
     * iteration unless it is the leftmost call.
     *
     * @param a the array to be sorted
     * @param low the index of the first element, inclusive, to be sorted
     * @param end the index of the last element for simple insertion sort
     * @param high the index of the last element, exclusive, to be sorted
     */
    private static void mixedInsertionSort(double[] a, int low, int end, int high) {
        if (end == high) {

            /*
             * Invoke simple insertion sort on tiny array.
             */
            for (int i; ++low < end; ) {
                double ai = a[i = low];

                while (ai < a[--i]) {
                    a[i + 1] = a[i];
                }
                a[i + 1] = ai;
            }
        } else {

            /*
             * Start with pin insertion sort on small part.
             *
             * Pin insertion sort is extended simple insertion sort.
             * The main idea of this sort is to put elements larger
             * than an element called pin to the end of array (the
             * proper area for such elements). It avoids expensive
             * movements of these elements through the whole array.
             */
            double pin = a[end];

            for (int i, p = high; ++low < end; ) {
                double ai = a[i = low];

                if (ai < a[i - 1]) { // Small element

                    /*
                     * Insert small element into sorted part.
                     */
                    a[i] = a[--i];

                    while (ai < a[--i]) {
                        a[i + 1] = a[i];
                    }
                    a[i + 1] = ai;

                } else if (p > i && ai > pin) { // Large element

                    /*
                     * Find element smaller than pin.
                     */
                    while (a[--p] > pin);

                    /*
                     * Swap it with large element.
                     */
                    if (p > i) {
                        ai = a[p];
                        a[p] = a[i];
                    }

                    /*
                     * Insert small element into sorted part.
                     */
                    while (ai < a[--i]) {
                        a[i + 1] = a[i];
                    }
                    a[i + 1] = ai;
                }
            }

            /*
             * Continue with pair insertion sort on remain part.
             */
            for (int i; low < high; ++low) {
                double a1 = a[i = low], a2 = a[++low];

                /*
                 * Insert two elements per iteration: at first, insert the
                 * larger element and then insert the smaller element, but
                 * from the position where the larger element was inserted.
                 */
                if (a1 > a2) {

                    while (a1 < a[--i]) {
                        a[i + 2] = a[i];
                    }
                    a[++i + 1] = a1;

                    while (a2 < a[--i]) {
                        a[i + 1] = a[i];
                    }
                    a[i + 1] = a2;

                } else if (a1 < a[i - 1]) {

                    while (a2 < a[--i]) {
                        a[i + 2] = a[i];
                    }
                    a[++i + 1] = a2;

                    while (a1 < a[--i]) {
                        a[i + 1] = a[i];
                    }
                    a[i + 1] = a1;
                }
            }
        }
    }

    /**
     * Sorts the specified range of the array using insertion sort.
     *
     * @param a the array to be sorted
     * @param low the index of the first element, inclusive, to be sorted
     * @param high the index of the last element, exclusive, to be sorted
     */
    private static void insertionSort(double[] a, int low, int high) {
        for (int i, k = low; ++k < high; ) {
            double ai = a[i = k];

            if (ai < a[i - 1]) {
                while (--i >= low && ai < a[i]) {
                    a[i + 1] = a[i];
                }
                a[i + 1] = ai;
            }
        }
    }

    /**
     * Sorts the specified range of the array using heap sort.
     *
     * @param a the array to be sorted
     * @param low the index of the first element, inclusive, to be sorted
     * @param high the index of the last element, exclusive, to be sorted
     */
    private static void heapSort(double[] a, int low, int high) {
        for (int k = (low + high) >>> 1; k > low; ) {
            pushDown(a, --k, a[k], low, high);
        }
        while (--high > low) {
            double max = a[low];
            pushDown(a, low, a[high], low, high);
            a[high] = max;
        }
    }

    /**
     * Pushes specified element down during heap sort.
     *
     * @param a the given array
     * @param p the start index
     * @param value the given element
     * @param low the index of the first element, inclusive, to be sorted
     * @param high the index of the last element, exclusive, to be sorted
     */
    private static void pushDown(double[] a, int p, double value, int low, int high) {
        for (int k ;; a[p] = a[p = k]) {
            k = (p << 1) - low + 2; // Index of the right child

            if (k > high) {
                break;
            }
            if (k == high || a[k] < a[k - 1]) {
                --k;
            }
            if (a[k] <= value) {
                break;
            }
        }
        a[p] = value;
    }

    /**
     * Tries to sort the specified range of the array.
     *
     * @param sorter parallel context
     * @param a the array to be sorted
     * @param low the index of the first element to be sorted
     * @param size the array size
     * @return true if finally sorted, false otherwise
     */
    private static boolean tryMergeRuns(Sorter sorter, double[] a, int low, int size) {

        /*
         * The run array is constructed only if initial runs are
         * long enough to continue, run[i] then holds start index
         * of the i-th sequence of elements in non-descending order.
         */
        int[] run = null;
        int high = low + size;
        int count = 1, last = low;

        /*
         * Identify all possible runs.
         */
        for (int k = low + 1; k < high; ) {

            /*
             * Find the end index of the current run.
             */
            if (a[k - 1] < a[k]) {

                // Identify ascending sequence
                while (++k < high && a[k - 1] <= a[k]);

            } else if (a[k - 1] > a[k]) {

                // Identify descending sequence
                while (++k < high && a[k - 1] >= a[k]);

                // Reverse into ascending order
                for (int i = last - 1, j = k; ++i < --j && a[i] > a[j]; ) {
                    double ai = a[i]; a[i] = a[j]; a[j] = ai;
                }
            } else { // Identify constant sequence
                for (double ak = a[k]; ++k < high && ak == a[k]; );

                if (k < high) {
                    continue;
                }
            }

            /*
             * Check special cases.
             */
            if (run == null) {
                if (k == high) {

                    /*
                     * The array is monotonous sequence,
                     * and therefore already sorted.
                     */
                    return true;
                }

                if (k - low < MIN_FIRST_RUN_SIZE) {

                    /*
                     * The first run is too small
                     * to proceed with scanning.
                     */
                    return false;
                }

                run = new int[((size >> 10) | 0x7F) & 0x3FF];
                run[0] = low;

            } else if (a[last - 1] > a[last]) {

                if (count > (k - low) >> MIN_FIRST_RUNS_FACTOR) {

                    /*
                     * The first runs are not long
                     * enough to continue scanning.
                     */
                    return false;
                }

                if (++count == MAX_RUN_CAPACITY) {

                    /*
                     * Array is not highly structured.
                     */
                    return false;
                }

                if (count == run.length) {

                    /*
                     * Increase capacity of index array.
                     */
                    run = Arrays.copyOf(run, count << 1);
                }
            }
            run[count] = (last = k);
        }

        /*
         * Merge runs of highly structured array.
         */
        if (count > 1) {
            double[] b; int offset = low;

            if (sorter == null || (b = (double[]) sorter.b) == null) {
                b = new double[size];
            } else {
                offset = sorter.offset;
            }
            mergeRuns(a, b, offset, 1, sorter != null, run, 0, count);
        }
        return true;
    }

    /**
     * Merges the specified runs.
     *
     * @param a the source array
     * @param b the temporary buffer used in merging
     * @param offset the start index in the source, inclusive
     * @param aim specifies merging: to source ( > 0), buffer ( < 0) or any ( == 0)
     * @param parallel indicates whether merging is performed in parallel
     * @param run the start indexes of the runs, inclusive
     * @param lo the start index of the first run, inclusive
     * @param hi the start index of the last run, inclusive
     * @return the destination where runs are merged
     */
    private static double[] mergeRuns(double[] a, double[] b, int offset,
            int aim, boolean parallel, int[] run, int lo, int hi) {

        if (hi - lo == 1) {
            if (aim >= 0) {
                return a;
            }
            for (int i = run[hi], j = i - offset, low = run[lo]; i > low;
                b[--j] = a[--i]
            );
            return b;
        }

        /*
         * Split into approximately equal parts.
         */
        int mi = lo, rmi = (run[lo] + run[hi]) >>> 1;
        while (run[++mi + 1] <= rmi);

        /*
         * Merge the left and right parts.
         */
        double[] a1, a2;

        if (parallel && hi - lo > MIN_RUN_COUNT) {
            RunMerger merger = new RunMerger(a, b, offset, 0, run, mi, hi).forkMe();
            a1 = mergeRuns(a, b, offset, -aim, true, run, lo, mi);
            a2 = (double[]) merger.getDestination();
        } else {
            a1 = mergeRuns(a, b, offset, -aim, false, run, lo, mi);
            a2 = mergeRuns(a, b, offset,    0, false, run, mi, hi);
        }

        double[] dst = a1 == a ? b : a;

        int k   = a1 == a ? run[lo] - offset : run[lo];
        int lo1 = a1 == b ? run[lo] - offset : run[lo];
        int hi1 = a1 == b ? run[mi] - offset : run[mi];
        int lo2 = a2 == b ? run[mi] - offset : run[mi];
        int hi2 = a2 == b ? run[hi] - offset : run[hi];

        if (parallel) {
            new Merger(null, dst, k, a1, lo1, hi1, a2, lo2, hi2).invoke();
        } else {
            mergeParts(null, dst, k, a1, lo1, hi1, a2, lo2, hi2);
        }
        return dst;
    }

    /**
     * Merges the sorted parts.
     *
     * @param merger parallel context
     * @param dst the destination where parts are merged
     * @param k the start index of the destination, inclusive
     * @param a1 the first part
     * @param lo1 the start index of the first part, inclusive
     * @param hi1 the end index of the first part, exclusive
     * @param a2 the second part
     * @param lo2 the start index of the second part, inclusive
     * @param hi2 the end index of the second part, exclusive
     */
    private static void mergeParts(Merger merger, double[] dst, int k,
            double[] a1, int lo1, int hi1, double[] a2, int lo2, int hi2) {

        if (merger != null && a1 == a2) {

            while (true) {

                /*
                 * The first part must be larger.
                 */
                if (hi1 - lo1 < hi2 - lo2) {
                    int lo = lo1; lo1 = lo2; lo2 = lo;
                    int hi = hi1; hi1 = hi2; hi2 = hi;
                }

                /*
                 * Small parts will be merged sequentially.
                 */
                if (hi1 - lo1 < MIN_PARALLEL_MERGE_PARTS_SIZE) {
                    break;
                }

                /*
                 * Find the median of the larger part.
                 */
                int mi1 = (lo1 + hi1) >>> 1;
                double key = a1[mi1];
                int mi2 = hi2;

                /*
                 * Partition the smaller part.
                 */
                for (int loo = lo2; loo < mi2; ) {
                    int t = (loo + mi2) >>> 1;

                    if (key > a2[t]) {
                        loo = t + 1;
                    } else {
                        mi2 = t;
                    }
                }

                int d = mi2 - lo2 + mi1 - lo1;

                /*
                 * Merge the right sub-parts in parallel.
                 */
                merger.forkMerger(dst, k + d, a1, mi1, hi1, a2, mi2, hi2);

                /*
                 * Process the sub-left parts.
                 */
                hi1 = mi1;
                hi2 = mi2;
            }
        }

        /*
         * Merge small parts sequentially.
         */
        while (lo1 < hi1 && lo2 < hi2) {
            dst[k++] = a1[lo1] < a2[lo2] ? a1[lo1++] : a2[lo2++];
        }
        if (dst != a1 || k < lo1) {
            while (lo1 < hi1) {
                dst[k++] = a1[lo1++];
            }
        }
        if (dst != a2 || k < lo2) {
            while (lo2 < hi2) {
                dst[k++] = a2[lo2++];
            }
        }
    }

// [class]

    /**
     * This class implements parallel sorting.
     */
    private static final class Sorter extends CountedCompleter<Void> {
        private static final long serialVersionUID = 20180818L;
        private final Object a, b;
        private final int low, size, offset, depth;

        private Sorter(CountedCompleter<?> parent,
                Object a, Object b, int low, int size, int offset, int depth) {
            super(parent);
            this.a = a;
            this.b = b;
            this.low = low;
            this.size = size;
            this.offset = offset;
            this.depth = depth;
        }

        @Override
        public final void compute() {
            if (depth < 0) {
                setPendingCount(2);
                int half = size >> 1;
                new Sorter(this, b, a, low, half, offset, depth + 1).fork();
                new Sorter(this, b, a, low + half, size - half, offset, depth + 1).compute();
            } else {
                if (a instanceof int[]) {
                    sort(this, (int[]) a, depth, low, low + size);
                } else if (a instanceof long[]) {
                    sort(this, (long[]) a, depth, low, low + size);
                } else if (a instanceof float[]) {
                    sort(this, (float[]) a, depth, low, low + size);
                } else if (a instanceof double[]) {
                    sort(this, (double[]) a, depth, low, low + size);
                } else {
                    throw new IllegalArgumentException(
                        "Unknown type of array: " + a.getClass().getName());
                }
            }
            tryComplete();
        }

        @Override
        public final void onCompletion(CountedCompleter<?> caller) {
            if (depth < 0) {
                int mi = low + (size >> 1);
                boolean src = (depth & 1) == 0;

                new Merger(null,
                    a,
                    src ? low : low - offset,
                    b,
                    src ? low - offset : low,
                    src ? mi - offset : mi,
                    b,
                    src ? mi - offset : mi,
                    src ? low + size - offset : low + size
                ).invoke();
            }
        }

        private void forkSorter(int depth, int low, int high) {
            addToPendingCount(1);
            Object a = this.a; // Use local variable for performance
            new Sorter(this, a, b, low, high - low, offset, depth).fork();
        }
    }

    /**
     * This class implements parallel merging.
     */
    private static final class Merger extends CountedCompleter<Void> {
        private static final long serialVersionUID = 20180818L;
        private final Object dst, a1, a2;
        private final int k, lo1, hi1, lo2, hi2;

        private Merger(CountedCompleter<?> parent, Object dst, int k,
                Object a1, int lo1, int hi1, Object a2, int lo2, int hi2) {
            super(parent);
            this.dst = dst;
            this.k = k;
            this.a1 = a1;
            this.lo1 = lo1;
            this.hi1 = hi1;
            this.a2 = a2;
            this.lo2 = lo2;
            this.hi2 = hi2;
        }

        @Override
        public final void compute() {
            if (dst instanceof int[]) {
                mergeParts(this, (int[]) dst, k,
                    (int[]) a1, lo1, hi1, (int[]) a2, lo2, hi2);
            } else if (dst instanceof long[]) {
                mergeParts(this, (long[]) dst, k,
                    (long[]) a1, lo1, hi1, (long[]) a2, lo2, hi2);
            } else if (dst instanceof float[]) {
                mergeParts(this, (float[]) dst, k,
                    (float[]) a1, lo1, hi1, (float[]) a2, lo2, hi2);
            } else if (dst instanceof double[]) {
                mergeParts(this, (double[]) dst, k,
                    (double[]) a1, lo1, hi1, (double[]) a2, lo2, hi2);
            } else {
                throw new IllegalArgumentException(
                    "Unknown type of array: " + dst.getClass().getName());
            }
            propagateCompletion();
        }

        private void forkMerger(Object dst, int k,
                Object a1, int lo1, int hi1, Object a2, int lo2, int hi2) {
            addToPendingCount(1);
            new Merger(this, dst, k, a1, lo1, hi1, a2, lo2, hi2).fork();
        }
    }

    /**
     * This class implements parallel merging of runs.
     */
    private static final class RunMerger extends RecursiveTask<Object> {
        private static final long serialVersionUID = 20180818L;
        private final Object a, b;
        private final int[] run;
        private final int offset, aim, lo, hi;

        private RunMerger(Object a, Object b, int offset,
                int aim, int[] run, int lo, int hi) {
            this.a = a;
            this.b = b;
            this.offset = offset;
            this.aim = aim;
            this.run = run;
            this.lo = lo;
            this.hi = hi;
        }

        @Override
        protected final Object compute() {
            if (a instanceof int[]) {
                return mergeRuns((int[]) a, (int[]) b, offset, aim, true, run, lo, hi);
            }
            if (a instanceof long[]) {
                return mergeRuns((long[]) a, (long[]) b, offset, aim, true, run, lo, hi);
            }
            if (a instanceof float[]) {
                return mergeRuns((float[]) a, (float[]) b, offset, aim, true, run, lo, hi);
            }
            if (a instanceof double[]) {
                return mergeRuns((double[]) a, (double[]) b, offset, aim, true, run, lo, hi);
            }
            throw new IllegalArgumentException(
                "Unknown type of array: " + a.getClass().getName());
        }

        private RunMerger forkMe() {
            fork();
            return this;
        }

        private Object getDestination() {
            join();
            return getRawResult();
        }
    }
}
