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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <string>
#include <sstream>
#include <complex>
#include <vector>
#include <list>
#include <queue>
#include <deque>
#include <stack>
#include <map>
#include <set>
#include <iomanip>
#include <limits>
using namespace std;
// 素因数分解
typedef pair<long long, long long> P;
vector<P> prime_factorize(long long n)
{
vector<P> res;
for (long long p = 2; p * p <= n; ++p)
{
if (n % p != 0)
continue;
int num = 0;
while (n % p == 0)
{
++num;
n /= p;
}
res.push_back(make_pair(p, num));
}
if (n != 1)
res.push_back(make_pair(n, 1));
return res;
}
// 因数分解
long long gcd(long long x, long long y)
{
  if (y == 0)
  {
    return x;
  }
  else
  {
    return gcd(y, x % y);
  }
}
const int MOD = 1000000007;
typedef long long int ll;
#define EPS (1e-7)
#define INF (1e9)
#define PI (acos(-1))
#define rep(i, n) for (int i = 0; i < (n); i++)

int main()
{
    double L;
    cin >> L;
    L /= 3.0;
    double ans = pow(L, 3);
    cout <<fixed<<setprecision(10)<< ans << endl;
    return 0;
}