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 Python (3.4.3) (judge verdict: Runtime Error)

#include <bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int i=0;i<n;i++)
#define mfill(x,y) memset(x,y,sizeof(x))
#define all(v) v.begin(), v.end()
#define in(x,y,h,w) if(0<=x&&x<h&&0<=y&&y<w)
#define y0 y12345
#define y1 y54321

#ifdef LOCAL
#define debug(...) fprintf(stderr,__VA_ARGS__)
#else
#define debug(...) 42
#endif

using ul = unsigned long;
using ll = long long;
using P = pair<int, int>;
using vint = vector<int>;using vvint = vector<vector<int>>;
using vll = vector<ll>;using vvll = vector<vector<ll>>;
template<class T>bool chmax(T &a, const T &b){if(a<b){a=b;return 1;}return 0;}
template<class T>bool chmin(T &a, const T &b){if(a>b){a=b;return 1;}return 0;}
template <class T>void initvv(vector<vector<T> > &v, int a, int b, const T &t = T()){
    v.assign(a, vector<T>(b, t));
}
template <class T>T gcd(T &a, T &b){if(a<b){swap(a,b);} T r = a%b; while(r!=0){a=b;b=r;r=a%b;} return b;}

vint dx = {-1, 0, 1, 0}, dy = {0, -1, 0, 1};
vint dx8 = {-1, -1, -1, 0, 1, 1, 1, 0}, dy8 = {-1, 0, 1, 1, 1, 0, -1, -1};

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    debug("debug test\n");
    int n,m;cin>>n;
    double l = (double)n/3.0;
    printf("%.10f\n", l*l*l);

    return 0;
}