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;

        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;

	}
}

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"},

    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);
    }

            final BigInteger firstBigInt = BigInteger.valueOf(first);
            BigInteger completeBigInt = lastBigInt.add(firstBigInt.shiftLeft(32));
            completeBigInt = completeBigInt.divide(BigInteger.valueOf(10000L));
            completeBigInt = completeBigInt.add(BigInteger.valueOf(-SmbConstants.MILLISECONDS_BETWEEN_1970_AND_1601));
            date = new Date(completeBigInt.longValue());
        }

- **MinIO creates erasure-coding sets of _2_ to _16_ drives per set.  The number of drives you provide in total must be a multiple of one of those numbers.**
- **MinIO chooses the largest EC set size which divides into the total number of drives or total number of nodes given - making sure to keep the uniform distribution i.e each node participates equal number of drives per set**.

        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);
        }
      }
    }
  }

    }

    @Test
    void testUTimeReadWrite() {
        byte[] buffer = new byte[4];
        long time = System.currentTimeMillis();
        // UTime is seconds since epoch, so divide by 1000
        long unixTime = time / 1000L;

        // Mocking date for timezone consistency
        TimeZone original = TimeZone.getDefault();
        try {

	return ep, nil
}

// GetAllSets - parses all ellipses input arguments, expands them into
// corresponding list of endpoints chunked evenly in accordance with a
// specific set size.
// For example: {1...64} is divided into 4 sets each of size 16.
// This applies to even distributed setup syntax as well.
func GetAllSets(setDriveCount uint64, args ...string) ([][]string, error) {
	var setArgs [][]string
	if !ellipses.HasEllipses(args...) {

      // Avoid delegating to this(long[]), since AtomicLongArray(long[]) will clone its input and
      // thus double memory usage.
      this.data =
          new AtomicLongArray(Ints.checkedCast(LongMath.divide(bits, 64, RoundingMode.CEILING)));
      this.bitCount = LongAddables.create();
    }

    // Used by serialization
    LockFreeBitArray(long[] data) {
      checkArgument(data.length > 0, "data length is zero!");