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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
#define ll long long 
#define endl "\n"
#define pb push_back
#define f(i,n) for(i=0;i<n;i++)
#define F(i,a,b) for(i=a;a<=b;i++)
#define arr(a,n) for( i=0;i<n;i++)cin>>a[i];
#define fi first
#define se second
#define mp make_pair
#define mod 1000000007
#define YES cout<<"YES"<<endl;
#define Yes cout<<"Yes"<<endl;
#define NO cout<<"NO"<<endl;
#define No cout<<"No"<<endl;
#define yes cout<<"yes"<<endl;
#define no cout<<"no"<<endl;
#define vi vector<ll>
#define ed end()
#define bg begin()
#define sz size()
#define ln length()
#define s() sort(a,a+n);
#define sr() sort(a,a+n,greater<ll>());
#define v()  sort(v.begin(),v.end());
#define vr() sort(v.begin(),v.end(),greater<ll>());
#define mod 1000000007
#define fast() ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)
ll gcd(ll a, ll b){if(!b)return a;return gcd(b, a % b);}
ll power(ll x,ll y,ll p){ll res=1;x%=p;while(y>0){if(y&1)res=(res*x)%p;y=y>>1;x=(x*x)%p;}return res;}
int main() {
   /* #ifndef ONLINE_JUDGE
        // for getting input from input.txt
        freopen("input.txt", "r", stdin);
        // for writing output to output.txt
        freopen("output.txt", "w", stdout);
    #endif*/
        /* 
        Ofcourse it's Hard.
        It's supposed to be Hard.
        If it's easy everyone would do it.
        HARD IS WHAT MAKES IT GREAT

        YESTERDAY U SAID TOMORROW

        SLOWLY BECOMING THE PERSON I 
        SHOULD HAVE BEEN A LONG TIME AGO

        SAME TASK CAN'T BE FOUND DIFFICULT TWICE

        BTBHWSITW

        SPRH TU KAB P

        CP IS ALL ABOUT THINKING 
        YOU HAVE MUCH MORE POTENTIAL THAN U THINK


        */

fast();
//ll t;cin>>t;while(t--)
{
   double n;
   cin>>n;
if(n==1)
{
    cout<<fixed<<setprecision(12)<<1<<endl;
}
else{
   double a=n/3;
   cerr<<a<<endl;
   double ans=a*a*a;
   cout<<fixed<<setprecision(12)<<ans<<endl;

  } 

   

}
return 0;
}