long result = 1;
      for (int i = n1 + 1; i <= n2; i++) {
        result *= i;
      }
      return BigInteger.valueOf(result);
    }

    /*
     * We want each multiplication to have both sides with approximately the same number of digits.
     * Currently, we just divide the range in half.
     */
    int mid = (n1 + n2) >>> 1;

      if (x.length() != (originalString.length() * count)) {
        throw new RuntimeException("Wrong length: " + x);
      }
    }
  }

  private static String oldRepeat(String string, int count) {
    // If this multiplication overflows, a NegativeArraySizeException or
    // OutOfMemoryError is not far behind
    int len = string.length();
    int size = len * count;
    char[] array = new char[size];
    for (int i = 0; i < size; i += len) {

      if (x.length() != (originalString.length() * count)) {
        throw new RuntimeException("Wrong length: " + x);
      }
    }
  }

  private static String oldRepeat(String string, int count) {
    // If this multiplication overflows, a NegativeArraySizeException or
    // OutOfMemoryError is not far behind
    int len = string.length();
    int size = len * count;
    char[] array = new char[size];
    for (int i = 0; i < size; i += len) {

      long result = 1;
      for (int i = n1 + 1; i <= n2; i++) {
        result *= i;
      }
      return BigInteger.valueOf(result);
    }

    /*
     * We want each multiplication to have both sides with approximately the same number of digits.
     * Currently, we just divide the range in half.
     */
    int mid = (n1 + n2) >>> 1;

    assertEquals(1, future.get(-1, SECONDS).intValue());
  }

  @J2ktIncompatible
  @GwtIncompatible // threads
  public void testOverflowTimeout() throws Exception {
    // First, sanity check that naive multiplication would really overflow to a negative number:
    long nanosPerSecond = NANOSECONDS.convert(1, SECONDS);
    assertThat(nanosPerSecond * Long.MAX_VALUE).isLessThan(0L);

      return true;
    }
  },
  /**
   * This strategy uses all 128 bits of {@link Hashing#murmur3_128} when hashing. It looks different
   * from the implementation in MURMUR128_MITZ_32 because we're avoiding the multiplication in the
   * loop and doing a (much simpler) += hash2. We're also changing the index to a positive number by
   * AND'ing with Long.MAX_VALUE instead of flipping the bits.
   */
  MURMUR128_MITZ_64() {
    @Override

    double ySumOfSquaresOfDeltas = yStats.sumOfSquaresOfDeltas();
    checkState(xSumOfSquaresOfDeltas > 0.0);
    checkState(ySumOfSquaresOfDeltas > 0.0);
    // The product of two positive numbers can be zero if the multiplication underflowed. We
    // force a positive value by effectively rounding up to MIN_VALUE.
    double productOfSumsOfSquaresOfDeltas =
        ensurePositive(xSumOfSquaresOfDeltas * ySumOfSquaresOfDeltas);

    double ySumOfSquaresOfDeltas = yStats().sumOfSquaresOfDeltas();
    checkState(xSumOfSquaresOfDeltas > 0.0);
    checkState(ySumOfSquaresOfDeltas > 0.0);
    // The product of two positive numbers can be zero if the multiplication underflowed. We
    // force a positive value by effectively rounding up to MIN_VALUE.
    double productOfSumsOfSquaresOfDeltas =
        ensurePositive(xSumOfSquaresOfDeltas * ySumOfSquaresOfDeltas);

    // Strip off 2s from this value.
    int shift = Long.numberOfTrailingZeros(product);
    product >>= shift;

    // Use floor(log2(num)) + 1 to prevent overflow of multiplication.
    int productBits = LongMath.log2(product, FLOOR) + 1;
    int bits = LongMath.log2(startingNumber, FLOOR) + 1;
    // Check for the next power of two boundary, to save us a CLZ operation.

			copy(s.buf, s.buf[s.start:s.end])
			s.end -= s.start
			s.start = 0
		}
		// Is the buffer full? If so, resize.
		if s.end == len(s.buf) {
			// Guarantee no overflow in the multiplication below.
			const maxInt = int(^uint(0) >> 1)
			if len(s.buf) >= s.maxTokenSize || len(s.buf) > maxInt/2 {
				s.setErr(ErrTooLong)
				return false
			}
			newSize := len(s.buf) * 2
			if newSize == 0 {