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package javatest.sort;

import java.util.Arrays;

class ComparableTimSort {
	/**
	 * This is the minimum sized sequence that will be merged.  Shorter
	 * sequences will be lengthened by calling binarySort.  If the entire
	 * array is less than this length, no merges will be performed.
	 *
	 * This constant should be a power of two.  It was 64 in Tim Peter's C
	 * implementation, but 32 was empirically determined to work better in
	 * this implementation.  In the unlikely event that you set this constant
	 * to be a number that's not a power of two, you'll need to change the
	 * {@link #minRunLength} computation.
	 *
	 * If you decrease this constant, you must change the stackLen
	 * computation in the TimSort constructor, or you risk an
	 * ArrayOutOfBounds exception.  See listsort.txt for a discussion
	 * of the minimum stack length required as a function of the length
	 * of the array being sorted and the minimum merge sequence length.
	 */
	private static final int MIN_MERGE = 32;

	/**
	 * The array being sorted.
	 */
	private final Object[] a;

	/**
	 * When we get into galloping mode, we stay there until both runs win less
	 * often than MIN_GALLOP consecutive times.
	 */
	private static final int  MIN_GALLOP = 7;

	/**
	 * This controls when we get *into* galloping mode.  It is initialized
	 * to MIN_GALLOP.  The mergeLo and mergeHi methods nudge it higher for
	 * random data, and lower for highly structured data.
	 */
	private int minGallop = MIN_GALLOP;

	/**
	 * Maximum initial size of tmp array, which is used for merging.  The array
	 * can grow to accommodate demand.
	 *
	 * Unlike Tim's original C version, we do not allocate this much storage
	 * when sorting smaller arrays.  This change was required for performance.
	 */
	private static final int INITIAL_TMP_STORAGE_LENGTH = 256;

	/**
	 * Temp storage for merges.
	 */
	private Object[] tmp;

	/**
	 * A stack of pending runs yet to be merged.  Run i starts at
	 * address base[i] and extends for len[i] elements.  It's always
	 * true (so long as the indices are in bounds) that:
	 *
	 *     runBase[i] + runLen[i] == runBase[i + 1]
	 *
	 * so we could cut the storage for this, but it's a minor amount,
	 * and keeping all the info explicit simplifies the code.
	 */
	private int stackSize = 0;  // Number of pending runs on stack
	private final int[] runBase;
	private final int[] runLen;

	/**
	 * Asserts have been placed in if-statements for performace. To enable them,
	 * set this field to true and enable them in VM with a command line flag.
	 * If you modify this class, please do test the asserts!
	 */
	private static final boolean DEBUG = false;

	/**
	 * Creates a TimSort instance to maintain the state of an ongoing sort.
	 *
	 * @param a the array to be sorted
	 */
	private ComparableTimSort(Object[] a) {
		this.a = a;

		// Allocate temp storage (which may be increased later if necessary)
		int len = a.length;
		@SuppressWarnings({"unchecked", "UnnecessaryLocalVariable"})
		Object[] newArray = new Object[len < 2 * INITIAL_TMP_STORAGE_LENGTH ?
			len >>> 1 : INITIAL_TMP_STORAGE_LENGTH];
		tmp = newArray;

        /*
         * Allocate runs-to-be-merged stack (which cannot be expanded).  The
         * stack length requirements are described in listsort.txt.  The C
         * version always uses the same stack length (85), but this was
         * measured to be too expensive when sorting "mid-sized" arrays (e.g.,
         * 100 elements) in Java.  Therefore, we use smaller (but sufficiently
         * large) stack lengths for smaller arrays.  The "magic numbers" in the
         * computation below must be changed if MIN_MERGE is decreased.  See
         * the MIN_MERGE declaration above for more information.
         */
		int stackLen = (len <    120  ?  5 :
			len <   1542  ? 10 :
				len < 119151  ? 19 : 40);
		runBase = new int[stackLen];
		runLen = new int[stackLen];
	}

    /*
     * The next two methods (which are package private and static) constitute
     * the entire API of this class.  Each of these methods obeys the contract
     * of the public method with the same signature in java.util.Arrays.
     */

	static void sort(Object[] a) {
		sort(a, 0, a.length);
	}

	static void sort(Object[] a, int lo, int hi) {
		//Arrays.checkStartAndEnd(a.length, lo, hi);
		int nRemaining  = hi - lo;
		if (nRemaining < 2)
			return;  // Arrays of size 0 and 1 are always sorted

		// If array is small, do a "mini-TimSort" with no merges
		if (nRemaining < MIN_MERGE) {
			int initRunLen = countRunAndMakeAscending(a, lo, hi);
			binarySort(a, lo, hi, lo + initRunLen);
			return;
		}

		/**
		 * March over the array once, left to right, finding natural runs,
		 * extending short natural runs to minRun elements, and merging runs
		 * to maintain stack invariant.
		 */
		ComparableTimSort ts = new ComparableTimSort(a);
		int minRun = minRunLength(nRemaining);
		do {
			// Identify next run
			int runLen = countRunAndMakeAscending(a, lo, hi);

			// If run is short, extend to min(minRun, nRemaining)
			if (runLen < minRun) {
				int force = nRemaining <= minRun ? nRemaining : minRun;
				binarySort(a, lo, lo + force, lo + runLen);
				runLen = force;
			}

			// Push run onto pending-run stack, and maybe merge
			ts.pushRun(lo, runLen);
			ts.mergeCollapse();

			// Advance to find next run
			lo += runLen;
			nRemaining -= runLen;
		} while (nRemaining != 0);

		// Merge all remaining runs to complete sort
		if (DEBUG) assert lo == hi;
		ts.mergeForceCollapse();
		if (DEBUG) assert ts.stackSize == 1;
	}

	/**
	 * Sorts the specified portion of the specified array using a binary
	 * insertion sort.  This is the best method for sorting small numbers
	 * of elements.  It requires O(n log n) compares, but O(n^2) data
	 * movement (worst case).
	 *
	 * If the initial part of the specified range is already sorted,
	 * this method can take advantage of it: the method assumes that the
	 * elements from index {@code lo}, inclusive, to {@code start},
	 * exclusive are already sorted.
	 *
	 * @param a the array in which a range is to be sorted
	 * @param lo the index of the first element in the range to be sorted
	 * @param hi the index after the last element in the range to be sorted
	 * @param start the index of the first element in the range that is
	 *        not already known to be sorted (@code lo <= start <= hi}
	 */
	@SuppressWarnings("fallthrough")
	private static void binarySort(Object[] a, int lo, int hi, int start) {
		if (DEBUG) assert lo <= start && start <= hi;
		if (start == lo)
			start++;
		for ( ; start < hi; start++) {
			System.out.printf("lo=%d, hi=%d, start=%d, array=%s\n", lo, hi, start, Arrays.toString(a));
			@SuppressWarnings("unchecked")
			Comparable pivot = (Comparable) a[start];

			// Set left (and right) to the index where a[start] (pivot) belongs
			int left = lo;
			int right = start;
			if (DEBUG) assert left <= right;
            /*
             * Invariants:
             *   pivot >= all in [lo, left).
             *   pivot <  all in [right, start).
             */
			while (left < right) {
				int mid = (left + right) >>> 1;
				if (pivot.compareTo(a[mid]) < 0)
					right = mid;
				else
					left = mid + 1;
			}
			if (DEBUG) assert left == right;

            /*
             * The invariants still hold: pivot >= all in [lo, left) and
             * pivot < all in [left, start), so pivot belongs at left.  Note
             * that if there are elements equal to pivot, left points to the
             * first slot after them -- that's why this sort is stable.
             * Slide elements over to make room to make room for pivot.
             */
			int n = start - left;  // The number of elements to move
			// Switch is just an optimization for arraycopy in default case
			switch(n) {
				case 2:
					System.out.println("[2]");
					a[left + 2] = a[left + 1];
				case 1:
					System.out.println("[1]");
					a[left + 1] = a[left];
					break;
				default:
					System.out.println("[default]");
					System.arraycopy(a, left, a, left + 1, n);
			}
			a[left] = pivot;
		}
	}

	/**
	 * Returns the length of the run beginning at the specified position in
	 * the specified array and reverses the run if it is descending (ensuring
	 * that the run will always be ascending when the method returns).
	 *
	 * A run is the longest ascending sequence with:
	 *
	 *    a[lo] <= a[lo + 1] <= a[lo + 2] <= ...
	 *
	 * or the longest descending sequence with:
	 *
	 *    a[lo] >  a[lo + 1] >  a[lo + 2] >  ...
	 *
	 * For its intended use in a stable mergesort, the strictness of the
	 * definition of "descending" is needed so that the call can safely
	 * reverse a descending sequence without violating stability.
	 *
	 * @param a the array in which a run is to be counted and possibly reversed
	 * @param lo index of the first element in the run
	 * @param hi index after the last element that may be contained in the run.
	It is required that @code{lo < hi}.
	 * @return  the length of the run beginning at the specified position in
	 *          the specified array
	 */
	@SuppressWarnings("unchecked")
	private static int countRunAndMakeAscending(Object[] a, int lo, int hi) {
		if (DEBUG) assert lo < hi;
		int runHi = lo + 1;
		if (runHi == hi)
			return 1;

		// Find end of run, and reverse range if descending
		if (((Comparable) a[runHi++]).compareTo(a[lo]) < 0) { // Descending
			while(runHi < hi && ((Comparable) a[runHi]).compareTo(a[runHi - 1]) < 0)
				runHi++;
			reverseRange(a, lo, runHi);
		} else {                              // Ascending
			while (runHi < hi && ((Comparable) a[runHi]).compareTo(a[runHi - 1]) >= 0)
				runHi++;
		}

		return runHi - lo;
	}

	/**
	 * Reverse the specified range of the specified array.
	 *
	 * @param a the array in which a range is to be reversed
	 * @param lo the index of the first element in the range to be reversed
	 * @param hi the index after the last element in the range to be reversed
	 */
	private static void reverseRange(Object[] a, int lo, int hi) {
		hi--;
		while (lo < hi) {
			Object t = a[lo];
			a[lo++] = a[hi];
			a[hi--] = t;
		}
	}

	/**
	 * Returns the minimum acceptable run length for an array of the specified
	 * length. Natural runs shorter than this will be extended with
	 * {@link #binarySort}.
	 *
	 * Roughly speaking, the computation is:
	 *
	 *  If n < MIN_MERGE, return n (it's too small to bother with fancy stuff).
	 *  Else if n is an exact power of 2, return MIN_MERGE/2.
	 *  Else return an int k, MIN_MERGE/2 <= k <= MIN_MERGE, such that n/k
	 *   is close to, but strictly less than, an exact power of 2.
	 *
	 * For the rationale, see listsort.txt.
	 *
	 * @param n the length of the array to be sorted
	 * @return the length of the minimum run to be merged
	 */
	private static int minRunLength(int n) {
		if (DEBUG) assert n >= 0;
		int r = 0;      // Becomes 1 if any 1 bits are shifted off
		while (n >= MIN_MERGE) {
			r |= (n & 1);
			n >>= 1;
		}
		return n + r;
	}

	/**
	 * Pushes the specified run onto the pending-run stack.
	 *
	 * @param runBase index of the first element in the run
	 * @param runLen  the number of elements in the run
	 */
	private void pushRun(int runBase, int runLen) {
		this.runBase[stackSize] = runBase;
		this.runLen[stackSize] = runLen;
		stackSize++;
	}

	/**
	 * Examines the stack of runs waiting to be merged and merges adjacent runs
	 * until the stack invariants are reestablished:
	 *
	 *     1. runLen[i - 3] > runLen[i - 2] + runLen[i - 1]
	 *     2. runLen[i - 2] > runLen[i - 1]
	 *
	 * This method is called each time a new run is pushed onto the stack,
	 * so the invariants are guaranteed to hold for i < stackSize upon
	 * entry to the method.
	 */
	private void mergeCollapse() {
		while (stackSize > 1) {
			int n = stackSize - 2;
			if (n > 0 && runLen[n-1] <= runLen[n] + runLen[n+1]) {
				if (runLen[n - 1] < runLen[n + 1])
					n--;
				mergeAt(n);
			} else if (runLen[n] <= runLen[n + 1]) {
				mergeAt(n);
			} else {
				break; // Invariant is established
			}
		}
	}

	/**
	 * Merges all runs on the stack until only one remains.  This method is
	 * called once, to complete the sort.
	 */
	private void mergeForceCollapse() {
		while (stackSize > 1) {
			int n = stackSize - 2;
			if (n > 0 && runLen[n - 1] < runLen[n + 1])
				n--;
			mergeAt(n);
		}
	}

	/**
	 * Merges the two runs at stack indices i and i+1.  Run i must be
	 * the penultimate or antepenultimate run on the stack.  In other words,
	 * i must be equal to stackSize-2 or stackSize-3.
	 *
	 * @param i stack index of the first of the two runs to merge
	 */
	@SuppressWarnings("unchecked")
	private void mergeAt(int i) {
		if (DEBUG) assert stackSize >= 2;
		if (DEBUG) assert i >= 0;
		if (DEBUG) assert i == stackSize - 2 || i == stackSize - 3;

		int base1 = runBase[i];
		int len1 = runLen[i];
		int base2 = runBase[i + 1];
		int len2 = runLen[i + 1];
		if (DEBUG) assert len1 > 0 && len2 > 0;
		if (DEBUG) assert base1 + len1 == base2;

        /*
         * Record the length of the combined runs; if i is the 3rd-last
         * run now, also slide over the last run (which isn't involved
         * in this merge).  The current run (i+1) goes away in any case.
         */
		runLen[i] = len1 + len2;
		if (i == stackSize - 3) {
			runBase[i + 1] = runBase[i + 2];
			runLen[i + 1] = runLen[i + 2];
		}
		stackSize--;

        /*
         * Find where the first element of run2 goes in run1. Prior elements
         * in run1 can be ignored (because they're already in place).
         */
		int k = gallopRight((Comparable) a[base2], a, base1, len1, 0);
		if (DEBUG) assert k >= 0;
		base1 += k;
		len1 -= k;
		if (len1 == 0)
			return;

        /*
         * Find where the last element of run1 goes in run2. Subsequent elements
         * in run2 can be ignored (because they're already in place).
         */
		len2 = gallopLeft((Comparable) a[base1 + len1 - 1], a,
			base2, len2, len2 - 1);
		if (DEBUG) assert len2 >= 0;
		if (len2 == 0)
			return;

		// Merge remaining runs, using tmp array with min(len1, len2) elements
		if (len1 <= len2)
			mergeLo(base1, len1, base2, len2);
		else
			mergeHi(base1, len1, base2, len2);
	}

	/**
	 * Locates the position at which to insert the specified key into the
	 * specified sorted range; if the range contains an element equal to key,
	 * returns the index of the leftmost equal element.
	 *
	 * @param key the key whose insertion point to search for
	 * @param a the array in which to search
	 * @param base the index of the first element in the range
	 * @param len the length of the range; must be > 0
	 * @param hint the index at which to begin the search, 0 <= hint < n.
	 *     The closer hint is to the result, the faster this method will run.
	 * @return the int k,  0 <= k <= n such that a[b + k - 1] < key <= a[b + k],
	 *    pretending that a[b - 1] is minus infinity and a[b + n] is infinity.
	 *    In other words, key belongs at index b + k; or in other words,
	 *    the first k elements of a should precede key, and the last n - k
	 *    should follow it.
	 */
	private static int gallopLeft(Comparable key, Object[] a,
								  int base, int len, int hint) {
		if (DEBUG) assert len > 0 && hint >= 0 && hint < len;

		int lastOfs = 0;
		int ofs = 1;
		if (key.compareTo(a[base + hint]) > 0) {
			// Gallop right until a[base+hint+lastOfs] < key <= a[base+hint+ofs]
			int maxOfs = len - hint;
			while (ofs < maxOfs && key.compareTo(a[base + hint + ofs]) > 0) {
				lastOfs = ofs;
				ofs = (ofs << 1) + 1;
				if (ofs <= 0)   // int overflow
					ofs = maxOfs;
			}
			if (ofs > maxOfs)
				ofs = maxOfs;

			// Make offsets relative to base
			lastOfs += hint;
			ofs += hint;
		} else { // key <= a[base + hint]
			// Gallop left until a[base+hint-ofs] < key <= a[base+hint-lastOfs]
			final int maxOfs = hint + 1;
			while (ofs < maxOfs && key.compareTo(a[base + hint - ofs]) <= 0) {
				lastOfs = ofs;
				ofs = (ofs << 1) + 1;
				if (ofs <= 0)   // int overflow
					ofs = maxOfs;
			}
			if (ofs > maxOfs)
				ofs = maxOfs;

			// Make offsets relative to base
			int tmp = lastOfs;
			lastOfs = hint - ofs;
			ofs = hint - tmp;
		}
		if (DEBUG) assert -1 <= lastOfs && lastOfs < ofs && ofs <= len;

        /*
         * Now a[base+lastOfs] < key <= a[base+ofs], so key belongs somewhere
         * to the right of lastOfs but no farther right than ofs.  Do a binary
         * search, with invariant a[base + lastOfs - 1] < key <= a[base + ofs].
         */
		lastOfs++;
		while (lastOfs < ofs) {
			int m = lastOfs + ((ofs - lastOfs) >>> 1);

			if (key.compareTo(a[base + m]) > 0)
				lastOfs = m + 1;  // a[base + m] < key
			else
				ofs = m;          // key <= a[base + m]
		}
		if (DEBUG) assert lastOfs == ofs;    // so a[base + ofs - 1] < key <= a[base + ofs]
		return ofs;
	}

	/**
	 * Like gallopLeft, except that if the range contains an element equal to
	 * key, gallopRight returns the index after the rightmost equal element.
	 *
	 * @param key the key whose insertion point to search for
	 * @param a the array in which to search
	 * @param base the index of the first element in the range
	 * @param len the length of the range; must be > 0
	 * @param hint the index at which to begin the search, 0 <= hint < n.
	 *     The closer hint is to the result, the faster this method will run.
	 * @return the int k,  0 <= k <= n such that a[b + k - 1] <= key < a[b + k]
	 */
	private static int gallopRight(Comparable key, Object[] a,
								   int base, int len, int hint) {
		if (DEBUG) assert len > 0 && hint >= 0 && hint < len;

		int ofs = 1;
		int lastOfs = 0;
		if (key.compareTo(a[base + hint]) < 0) {
			// Gallop left until a[b+hint - ofs] <= key < a[b+hint - lastOfs]
			int maxOfs = hint + 1;
			while (ofs < maxOfs && key.compareTo(a[base + hint - ofs]) < 0) {
				lastOfs = ofs;
				ofs = (ofs << 1) + 1;
				if (ofs <= 0)   // int overflow
					ofs = maxOfs;
			}
			if (ofs > maxOfs)
				ofs = maxOfs;

			// Make offsets relative to b
			int tmp = lastOfs;
			lastOfs = hint - ofs;
			ofs = hint - tmp;
		} else { // a[b + hint] <= key
			// Gallop right until a[b+hint + lastOfs] <= key < a[b+hint + ofs]
			int maxOfs = len - hint;
			while (ofs < maxOfs && key.compareTo(a[base + hint + ofs]) >= 0) {
				lastOfs = ofs;
				ofs = (ofs << 1) + 1;
				if (ofs <= 0)   // int overflow
					ofs = maxOfs;
			}
			if (ofs > maxOfs)
				ofs = maxOfs;

			// Make offsets relative to b
			lastOfs += hint;
			ofs += hint;
		}
		if (DEBUG) assert -1 <= lastOfs && lastOfs < ofs && ofs <= len;

        /*
         * Now a[b + lastOfs] <= key < a[b + ofs], so key belongs somewhere to
         * the right of lastOfs but no farther right than ofs.  Do a binary
         * search, with invariant a[b + lastOfs - 1] <= key < a[b + ofs].
         */
		lastOfs++;
		while (lastOfs < ofs) {
			int m = lastOfs + ((ofs - lastOfs) >>> 1);

			if (key.compareTo(a[base + m]) < 0)
				ofs = m;          // key < a[b + m]
			else
				lastOfs = m + 1;  // a[b + m] <= key
		}
		if (DEBUG) assert lastOfs == ofs;    // so a[b + ofs - 1] <= key < a[b + ofs]
		return ofs;
	}

	/**
	 * Merges two adjacent runs in place, in a stable fashion.  The first
	 * element of the first run must be greater than the first element of the
	 * second run (a[base1] > a[base2]), and the last element of the first run
	 * (a[base1 + len1-1]) must be greater than all elements of the second run.
	 *
	 * For performance, this method should be called only when len1 <= len2;
	 * its twin, mergeHi should be called if len1 >= len2.  (Either method
	 * may be called if len1 == len2.)
	 *
	 * @param base1 index of first element in first run to be merged
	 * @param len1  length of first run to be merged (must be > 0)
	 * @param base2 index of first element in second run to be merged
	 *        (must be aBase + aLen)
	 * @param len2  length of second run to be merged (must be > 0)
	 */
	@SuppressWarnings("unchecked")
	private void mergeLo(int base1, int len1, int base2, int len2) {
		if (DEBUG) assert len1 > 0 && len2 > 0 && base1 + len1 == base2;

		// Copy first run into temp array
		Object[] a = this.a; // For performance
		Object[] tmp = ensureCapacity(len1);
		System.arraycopy(a, base1, tmp, 0, len1);

		int cursor1 = 0;       // Indexes into tmp array
		int cursor2 = base2;   // Indexes int a
		int dest = base1;      // Indexes int a

		// Move first element of second run and deal with degenerate cases
		a[dest++] = a[cursor2++];
		if (--len2 == 0) {
			System.arraycopy(tmp, cursor1, a, dest, len1);
			return;
		}
		if (len1 == 1) {
			System.arraycopy(a, cursor2, a, dest, len2);
			a[dest + len2] = tmp[cursor1]; // Last elt of run 1 to end of merge
			return;
		}

		int minGallop = this.minGallop;  // Use local variable for performance
		outer:
		while (true) {
			int count1 = 0; // Number of times in a row that first run won
			int count2 = 0; // Number of times in a row that second run won

            /*
             * Do the straightforward thing until (if ever) one run starts
             * winning consistently.
             */
			do {
				if (DEBUG) assert len1 > 1 && len2 > 0;
				if (((Comparable) a[cursor2]).compareTo(tmp[cursor1]) < 0) {
					a[dest++] = a[cursor2++];
					count2++;
					count1 = 0;
					if (--len2 == 0)
						break outer;
				} else {
					a[dest++] = tmp[cursor1++];
					count1++;
					count2 = 0;
					if (--len1 == 1)
						break outer;
				}
			} while ((count1 | count2) < minGallop);

            /*
             * One run is winning so consistently that galloping may be a
             * huge win. So try that, and continue galloping until (if ever)
             * neither run appears to be winning consistently anymore.
             */
			do {
				if (DEBUG) assert len1 > 1 && len2 > 0;
				count1 = gallopRight((Comparable) a[cursor2], tmp, cursor1, len1, 0);
				if (count1 != 0) {
					System.arraycopy(tmp, cursor1, a, dest, count1);
					dest += count1;
					cursor1 += count1;
					len1 -= count1;
					if (len1 <= 1)  // len1 == 1 || len1 == 0
						break outer;
				}
				a[dest++] = a[cursor2++];
				if (--len2 == 0)
					break outer;

				count2 = gallopLeft((Comparable) tmp[cursor1], a, cursor2, len2, 0);
				if (count2 != 0) {
					System.arraycopy(a, cursor2, a, dest, count2);
					dest += count2;
					cursor2 += count2;
					len2 -= count2;
					if (len2 == 0)
						break outer;
				}
				a[dest++] = tmp[cursor1++];
				if (--len1 == 1)
					break outer;
				minGallop--;
			} while (count1 >= MIN_GALLOP | count2 >= MIN_GALLOP);
			if (minGallop < 0)
				minGallop = 0;
			minGallop += 2;  // Penalize for leaving gallop mode
		}  // End of "outer" loop
		this.minGallop = minGallop < 1 ? 1 : minGallop;  // Write back to field

		if (len1 == 1) {
			if (DEBUG) assert len2 > 0;
			System.arraycopy(a, cursor2, a, dest, len2);
			a[dest + len2] = tmp[cursor1]; //  Last elt of run 1 to end of merge
		} else if (len1 == 0) {
			throw new IllegalArgumentException(
				"Comparison method violates its general contract!");
		} else {
			if (DEBUG) assert len2 == 0;
			if (DEBUG) assert len1 > 1;
			System.arraycopy(tmp, cursor1, a, dest, len1);
		}
	}

	/**
	 * Like mergeLo, except that this method should be called only if
	 * len1 >= len2; mergeLo should be called if len1 <= len2.  (Either method
	 * may be called if len1 == len2.)
	 *
	 * @param base1 index of first element in first run to be merged
	 * @param len1  length of first run to be merged (must be > 0)
	 * @param base2 index of first element in second run to be merged
	 *        (must be aBase + aLen)
	 * @param len2  length of second run to be merged (must be > 0)
	 */
	@SuppressWarnings("unchecked")
	private void mergeHi(int base1, int len1, int base2, int len2) {
		if (DEBUG) assert len1 > 0 && len2 > 0 && base1 + len1 == base2;

		// Copy second run into temp array
		Object[] a = this.a; // For performance
		Object[] tmp = ensureCapacity(len2);
		System.arraycopy(a, base2, tmp, 0, len2);

		int cursor1 = base1 + len1 - 1;  // Indexes into a
		int cursor2 = len2 - 1;          // Indexes into tmp array
		int dest = base2 + len2 - 1;     // Indexes into a

		// Move last element of first run and deal with degenerate cases
		a[dest--] = a[cursor1--];
		if (--len1 == 0) {
			System.arraycopy(tmp, 0, a, dest - (len2 - 1), len2);
			return;
		}
		if (len2 == 1) {
			dest -= len1;
			cursor1 -= len1;
			System.arraycopy(a, cursor1 + 1, a, dest + 1, len1);
			a[dest] = tmp[cursor2];
			return;
		}

		int minGallop = this.minGallop;  // Use local variable for performance
		outer:
		while (true) {
			int count1 = 0; // Number of times in a row that first run won
			int count2 = 0; // Number of times in a row that second run won

            /*
             * Do the straightforward thing until (if ever) one run
             * appears to win consistently.
             */
			do {
				if (DEBUG) assert len1 > 0 && len2 > 1;
				if (((Comparable) tmp[cursor2]).compareTo(a[cursor1]) < 0) {
					a[dest--] = a[cursor1--];
					count1++;
					count2 = 0;
					if (--len1 == 0)
						break outer;
				} else {
					a[dest--] = tmp[cursor2--];
					count2++;
					count1 = 0;
					if (--len2 == 1)
						break outer;
				}
			} while ((count1 | count2) < minGallop);

            /*
             * One run is winning so consistently that galloping may be a
             * huge win. So try that, and continue galloping until (if ever)
             * neither run appears to be winning consistently anymore.
             */
			do {
				if (DEBUG) assert len1 > 0 && len2 > 1;
				count1 = len1 - gallopRight((Comparable) tmp[cursor2], a, base1, len1, len1 - 1);
				if (count1 != 0) {
					dest -= count1;
					cursor1 -= count1;
					len1 -= count1;
					System.arraycopy(a, cursor1 + 1, a, dest + 1, count1);
					if (len1 == 0)
						break outer;
				}
				a[dest--] = tmp[cursor2--];
				if (--len2 == 1)
					break outer;

				count2 = len2 - gallopLeft((Comparable) a[cursor1], tmp, 0, len2, len2 - 1);
				if (count2 != 0) {
					dest -= count2;
					cursor2 -= count2;
					len2 -= count2;
					System.arraycopy(tmp, cursor2 + 1, a, dest + 1, count2);
					if (len2 <= 1)
						break outer; // len2 == 1 || len2 == 0
				}
				a[dest--] = a[cursor1--];
				if (--len1 == 0)
					break outer;
				minGallop--;
			} while (count1 >= MIN_GALLOP | count2 >= MIN_GALLOP);
			if (minGallop < 0)
				minGallop = 0;
			minGallop += 2;  // Penalize for leaving gallop mode
		}  // End of "outer" loop
		this.minGallop = minGallop < 1 ? 1 : minGallop;  // Write back to field

		if (len2 == 1) {
			if (DEBUG) assert len1 > 0;
			dest -= len1;
			cursor1 -= len1;
			System.arraycopy(a, cursor1 + 1, a, dest + 1, len1);
			a[dest] = tmp[cursor2];  // Move first elt of run2 to front of merge
		} else if (len2 == 0) {
			throw new IllegalArgumentException(
				"Comparison method violates its general contract!");
		} else {
			if (DEBUG) assert len1 == 0;
			if (DEBUG) assert len2 > 0;
			System.arraycopy(tmp, 0, a, dest - (len2 - 1), len2);
		}
	}

	/**
	 * Ensures that the external array tmp has at least the specified
	 * number of elements, increasing its size if necessary.  The size
	 * increases exponentially to ensure amortized linear time complexity.
	 *
	 * @param minCapacity the minimum required capacity of the tmp array
	 * @return tmp, whether or not it grew
	 */
	private Object[]  ensureCapacity(int minCapacity) {
		if (tmp.length < minCapacity) {
			// Compute smallest power of 2 > minCapacity
			int newSize = minCapacity;
			newSize |= newSize >> 1;
			newSize |= newSize >> 2;
			newSize |= newSize >> 4;
			newSize |= newSize >> 8;
			newSize |= newSize >> 16;
			newSize++;

			if (newSize < 0) // Not bloody likely!
				newSize = minCapacity;
			else
				newSize = Math.min(newSize, a.length >>> 1);

			@SuppressWarnings({"unchecked", "UnnecessaryLocalVariable"})
			Object[] newArray = new Object[newSize];
			tmp = newArray;
		}
		return tmp;
	}
}
    

    

    

            

    

            



    
    






    
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