} 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";

  // 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 {

      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;

  // 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 {

      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;

      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;
  }

        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"},