// to the given dividend, so that we don't have half of our divisions being
  // trivial because the divisor is bigger than the dividend.
  // Using remainder here does not give us a uniform distribution but it should
  // not have a big impact on the measurement.
  private static long randomDivisor(long dividend) {
    long r = randomSource.nextLong();
    if (dividend == -1) {
      return r;
    } else {

  // to the given dividend, so that we don't have half of our divisions being
  // trivial because the divisor is bigger than the dividend.
  // Using remainder here does not give us a uniform distribution but it should
  // not have a big impact on the measurement.
  private static long randomDivisor(long dividend) {
    long r = randomSource.nextLong();
    if (dividend == -1) {
      return r;
    } else {

    return fromLongBits(value * checkNotNull(val).value);
  }

  /**
   * Returns the result of dividing this by {@code val}.
   *
   * @since 14.0
   */
  public UnsignedLong dividedBy(UnsignedLong val) {
    return fromLongBits(UnsignedLongs.divide(value, checkNotNull(val).value));
  }

  /**
   * Returns this modulo {@code val}.
   *
   * @since 14.0
   */

    return fromIntBits(value * checkNotNull(val).value);
  }

  /**
   * Returns the result of dividing this by {@code val}.
   *
   * @throws ArithmeticException if {@code val} is zero
   * @since 14.0
   */
  public UnsignedInteger dividedBy(UnsignedInteger val) {
    return fromIntBits(UnsignedInts.divide(value, checkNotNull(val).value));
  }

  /**
   * Returns this mod {@code val}.
   *

        } else if (size.divide(ONE_GB_BI).compareTo(BigInteger.ZERO) > 0) {
            displaySize = new BigDecimal(size.divide(ONE_MB_BI)).divide(BigDecimal.valueOf(1000)) + "GB";
        } else if (size.divide(ONE_MB_BI).compareTo(BigInteger.ZERO) > 0) {
            displaySize = new BigDecimal(size.divide(ONE_KB_BI)).divide(BigDecimal.valueOf(1000)) + "MB";

## How are drives used for Erasure Code?

MinIO divides the drives you provide into erasure-coding sets of *2 to 16* drives.  Therefore, the number of drives you present must be a multiple of one of these numbers.  Each object is written to a single erasure-coding set.

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

  @Benchmark
  int divide(int reps) {
    int tmp = 0;
    for (int i = 0; i < reps; i++) {
      int j = i & ARRAY_MASK;
      tmp += BigIntegerMath.divide(nonzero1[j], nonzero2[j], mode).intValue();
    }
    return tmp;
  }

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

  }

  /**
   * 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 += 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 += BigIntegerMath.sqrt(positive[j], mode).intValue();
    }
    return tmp;
  }

  @Benchmark
  int divide(int reps) {
    int tmp = 0;
    for (int i = 0; i < reps; i++) {
      int j = i & ARRAY_MASK;
      tmp += BigIntegerMath.divide(nonzero1[j], nonzero2[j], mode).intValue();
    }
    return tmp;
  }

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