long result = 1;
      for (int i = n1 + 1; i <= n2; i++) {
        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));
  }

      long result = 1;
      for (int i = n1 + 1; i <= n2; i++) {
        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));
  }

   * returned from {@link #asMapOfRanges} will be different if there were existing entries which
   * connect to the given range and value.
   *
   * <p>Even if the input range is empty, if it is connected on both sides by ranges mapped to the
   * same value those two ranges will be coalesced.
   *
   * <p><b>Note:</b> coalescing requires calling {@code .equals()} on any connected values, which

   * returned from {@link #asMapOfRanges} will be different if there were existing entries which
   * connect to the given range and value.
   *
   * <p>Even if the input range is empty, if it is connected on both sides by ranges mapped to the
   * same value those two ranges will be coalesced.
   *
   * <p><b>Note:</b> coalescing requires calling {@code .equals()} on any connected values, which