Problem:
Score : 300 points

Problem Statement

Given is a positive integer L.
Find the maximum possible volume of a rectangular cuboid whose sum of the dimensions (not necessarily integers) is L.

Constraints

1 ≤ L ≤ 1000

L is an integer.

Input

Input is given from Standard Input in the following format:

L

Output

Print the maximum possible volume of a rectangular cuboid whose sum of the dimensions (not necessarily integers) is L.
Your output is considered correct if its absolute or relative error from our answer is at most 10^{-6}.

Sample Input 1

3

Sample Output 1

1.000000000000

For example, a rectangular cuboid whose dimensions are 0.8, 1, and 1.2 has a volume of 0.96.

On the other hand, if the dimensions are 1, 1, and 1, the volume of the rectangular cuboid is 1, which is greater.

Sample Input 2

999

Sample Output 2

36926037.000000000000

Solution in C# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;

namespace AtCoder_solve
{
    class Program
    {
        public const ulong mod = 1000000007;

        static void Main(string[] args)

        {

            double a = double.Parse(Console.ReadLine()) / 3.0;
            double ans = a * a * a;

                Console.WriteLine(ans);




        }

       
        static public long Input()
        {
            return long.Parse(Console.ReadLine());
        }

        static public string[] Input2()
        {
            return Console.ReadLine().Split();
        }

        static public ulong[] Input3()
        {
            string[] data = Console.ReadLine().Split();
            ulong[] res = new ulong[data.LongLength];
            for(long i = 0; i < data.LongLength; i++)
            {
                res[i] = ulong.Parse(data[i]);
            }
            return res;
        }
        public static long Choose(long n, long a)
        {
            long x = 1;
            long y = 1;
            for (long i = 0; i < a; i++)
            {
                x *= n - i;
                y *= i + 1;
            }
            return x / y;
        }
    }
}