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
   */

 x     y     x / y     x % y
 5     3       1         2
-5     3      -1        -2
 5    -3      -1         2
-5    -3       1        -2
</pre>

<p>
The one exception to this rule is that if the dividend <code>x</code> is
the most negative value for the int type of <code>x</code>, the quotient
<code>q = x / -1</code> is equal to <code>x</code> (and <code>r = 0</code>)

        } 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";

    a >>= aTwos; // divide out all 2s
    int bTwos = Long.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.

    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}.
   *

            byte[] testBuffer = new byte[8];
            ByteBuffer bb = ByteBuffer.wrap(testBuffer).order(ByteOrder.LITTLE_ENDIAN);
            bb.putShort((short) 0); // pathConsumed (will be divided by 2)
            bb.putShort((short) 0); // numReferrals
            bb.putShort((short) 0); // tflags
            bb.putShort((short) 0); // tflags high bytes (skipped in decode)

      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;

   * @param iterable the iterable to return a partitioned view of
   * @param size the desired size of each partition (the last may be smaller)
   * @return an iterable of unmodifiable lists containing the elements of {@code iterable} divided
   *     into partitions
   * @throws IllegalArgumentException if {@code size} is nonpositive
   */
  public static <T extends @Nullable Object> Iterable<List<T>> partition(
      Iterable<T> iterable, int size) {

   * @param iterable the iterable to return a partitioned view of
   * @param size the desired size of each partition (the last may be smaller)
   * @return an iterable of unmodifiable lists containing the elements of {@code iterable} divided
   *     into partitions
   * @throws IllegalArgumentException if {@code size} is nonpositive
   */
  public static <T extends @Nullable Object> Iterable<List<T>> partition(
      Iterable<T> iterable, int size) {

## 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.