new RoundToDoubleTester(BigDecimal.ZERO).setExpectation(0.0, RoundingMode.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)

  }

  /**
   * Binary search [sections] for [codePoint], looking at its top 14 bits.
   *
   * This binary searches over 4-byte entries, and so it needs to adjust binary search indices
   * in (by dividing by 4) and out (by multiplying by 4).
   */
  private fun findSectionsIndex(codePoint: Int): Int {
    val target = (codePoint and 0x1fff80) shr 7
    val offset =
      binarySearch(
        position = 0,

      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;
  }

Below is a list of data/parity drives and corresponding _approximate_ storage space usage on a 16 drive MinIO deployment. The field _storage
usage ratio_ is simply the drive space used by the file after erasure-encoding, divided by actual file size.

| Total Drives (N) | Data Drives (D) | Parity Drives (P) | Storage Usage Ratio |
|------------------|-----------------|-------------------|---------------------|

    }

    val roughSizeOfEachBucket = list.sumOf(toIntFunction) / expectedBucketNumber
    if (roughSizeOfEachBucket == 0) {
        // The elements in the list are so small that they can't even be divided into {expectedBucketNumber}.
        // For example, how do you split [0,0,0,0,0] into 3 buckets?
        // In this case, we simply put the elements into these buckets evenly.

	}
}

type badExprTest struct {
	input string
	error string // Empty means no error.
}

var badExprTests = []badExprTest{
	{"0/0", "division by zero"},
	{"3/0", "division by zero"},
	{"(1<<63)/0", "divide of value with high bit set"},
	{"3%0", "modulo by zero"},
	{"(1<<63)%0", "modulo of value with high bit set"},
	{"3<<-4", "negative left shift count"},
	{"3<<(1<<63)", "negative left shift count"},

        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;

        assertEquals(0x0010, response.getAttributes());
        // getLastWriteTime returns lastWriteTime (from readUTime) + serverTimeZoneOffset
        // readUTime multiplies the seconds value by 1000, and writeUTime divides milliseconds by 1000
        // So the round-trip should preserve the milliseconds value
        assertEquals(sampleTimeMillis + serverTimeZoneOffset, response.getLastWriteTime());
        assertEquals(1024, response.getSize());

        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;

    }

    private double singleQuantileFromSorted(int index, int scale, double[] dataset) {
      long numerator = (long) index * (dataset.length - 1);
      int positionFloor = (int) LongMath.divide(numerator, scale, RoundingMode.DOWN);
      int remainder = (int) (numerator - positionFloor * scale);
      if (remainder == 0) {
        return dataset[positionFloor];
      } else {