if (b != 0) {
          UnsignedInteger aUnsigned = UnsignedInteger.fromIntBits(a);
          UnsignedInteger bUnsigned = UnsignedInteger.fromIntBits(b);
          int expected = aUnsigned.bigIntegerValue().divide(bUnsigned.bigIntegerValue()).intValue();
          UnsignedInteger unsignedDiv = aUnsigned.dividedBy(bUnsigned);
          assertThat(unsignedDiv.intValue()).isEqualTo(expected);
        }
      }
    }
  }

    int bytesToCopy = bufferSize - buffer.position();
    for (int i = 0; i < bytesToCopy; i++) {
      buffer.put(readBuffer.get());
    }
    munch(); // buffer becomes empty here, since chunkSize divides bufferSize

    // Now process directly from the rest of the input buffer
    while (readBuffer.remaining() >= chunkSize) {
      process(readBuffer);
    }

    a >>= aTwos; // divide out all 2s
    int bTwos = Integer.numberOfTrailingZeros(b);
    b >>= bTwos; // divide out all 2s
    while (a != b) { // both a, b are odd
      // The key to the binary GCD algorithm is as follows:
      // Both a and b are odd. Assume a > b; then gcd(a - b, b) = gcd(a, b).
      // But in gcd(a - b, b), a - b is even and b is odd, so we can divide out powers of two.

      int start = index * size;
      int end = Math.min(start + size, list.size());
      return list.subList(start, end);
    }

    @Override
    public int size() {
      return IntMath.divide(list.size(), size, RoundingMode.CEILING);
    }

    @Override
    public boolean isEmpty() {
      return list.isEmpty();
    }
  }

      int j = i & ARRAY_MASK;
      tmp += IntMath.sqrt(positive[j], mode);
    }
    return tmp;
  }

  @Benchmark
  int divide(int reps) {
    int tmp = 0;
    for (int i = 0; i < reps; i++) {
      int j = i & ARRAY_MASK;
      tmp += IntMath.divide(ints[j], nonzero[j], mode);
    }
    return tmp;
  }

      int j = i & ARRAY_MASK;
      tmp += LongMath.sqrt(positive[j], mode);
    }
    return tmp;
  }

  @Benchmark
  int divide(int reps) {
    int tmp = 0;
    for (int i = 0; i < reps; i++) {
      int j = i & ARRAY_MASK;
      tmp += LongMath.divide(longs[j], nonzero[j], mode);
    }
    return tmp;
  }

    new RoundToDoubleTester(BigDecimal.ZERO).setExpectation(0.0, values()).test();
  }

  public void testRoundToDouble_oneThird() {
    new RoundToDoubleTester(
            BigDecimal.ONE.divide(BigDecimal.valueOf(3), new MathContext(50, HALF_EVEN)))
        .roundUnnecessaryShouldThrow()
        .setExpectation(0.33333333333333337, UP, CEILING)

      hexStrings[i] = UnsignedLongs.toString(longs[i], 16);
      prefixedHexStrings[i] = "0x" + hexStrings[i];
    }
  }

  @Benchmark
  long divide(int reps) {
    long tmp = 0;
    for (int i = 0; i < reps; i++) {
      int j = i & ARRAY_MASK;
      tmp += UnsignedLongs.divide(longs[j], divisors[j]);
    }
    return tmp;
  }

  @Benchmark
  long remainder(int reps) {
    long tmp = 0;

        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;
    return oldSlowFactorial(n1, mid).multiply(oldSlowFactorial(mid, n2));
  }

  @Benchmark
  int slowFactorial(int reps) {
    int tmp = 0;

    return Tables.transformValues(table, DIVIDE_BY_2);
  }

  @Override
  protected Map<String, Integer> makePopulatedMap() {
    Table<String, Character, Integer> table = HashBasedTable.create();
    table.put("one", 'a', 1);
    table.put("two", 'a', 2);
    table.put("three", 'a', 3);
    table.put("four", 'b', 4);
    return Tables.transformValues(table, DIVIDE_BY_2).column('a');
  }