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 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <math.h>
#include <stdbool.h>
#define LL long long
#define DB double
#define INF 1145141919810364364
#define PI 3.14159265358979
#define MIN(x,y) ((x)<(y)?(x):(y))
#define MAX(x,y) ((x)<(y)?(y):(x))
#define FOR(i,a,n) for(i=a;i<n;i++)
#define rep(i,n) FOR(i,0,n)
#define MOD 1000000007
//#define MOD 998244353
#define ERR 0.00000000000001
#define NUM 200010

void swap(LL* a,LL* b){LL tmp=*a;*a=*b;*b=tmp;}
void YN(int a){if(a==1)printf("Yes\n");if(a==0)printf("No\n");}
LL pwmn(LL a,LL n){LL ans=1;while(ans<a)ans*=n;return ans;}
LL GCD(LL a,LL b){LL c=1,tmp=MAX(a,b);b=MIN(a,b);a=tmp;while(c!=0){c=a%b;a=b;b=c;}return a;}
LL LCM(LL a,LL b){return a*b/GCD(a,b);}
int lcmp(const void* a,const void* b){if(*(LL*)a>*(LL*)b)return 1;if(*(LL*)a<*(LL*)b)return -1;return 0;}
int lcmpr(const void* a,const void* b){if(*(LL*)a>*(LL*)b)return -1;if(*(LL*)a<*(LL*)b)return 1;return 0;}
int ccmp(const void* a,const void* b){return *(char*)a-*(char*)b;}
int ccmpr(const void* a,const void* b){return *(char*)b-*(char*)a;}
int scmp(const void* a,const void* b){return strcmp((char*)a,(char*)b);}
int scmpr(const void* a,const void* b){return strcmp((char*)b,(char*)a);}
LL mod(LL a,LL m){if(a<0)return a%m+m;else return a%m;}
LL DIV(LL a,LL d){LL m=MOD,x=1,y=0,k;while(m){k=d/m;d-=k*m;swap(&m,&d);x-=k*y;swap(&x,&y);}return mod(a*mod(x,MOD),MOD);}
LL FAC(LL a){LL i,ans=1;FOR(i,1,a+1){ans*=i;if(MOD>0&&ans>MOD)ans%=MOD;}return ans;}
LL POW(LL a,LL n){LL ans=1;while(n>0){if(n&1)ans=ans*a%MOD;a=a*a%MOD;n>>=1;}return ans;}
LL fact[NUM],finv[NUM],inv[NUM];
void comi(){LL i;fact[0]=fact[1]=1;finv[0]=finv[1]=1;inv[1]=1;FOR(i,2,NUM){fact[i]=fact[i-1]*i%MOD;inv[i]=MOD-inv[MOD%i]*(MOD/i)%MOD;finv[i]=finv[i-1]*inv[i]%MOD;}}
LL com(LL n,LL k){if(n<k||n<0||k<0)return 0;return fact[n]*(finv[k]*finv[n-k]%MOD)%MOD;}

int main(){
  LL L,i,ans=0;
  DB l;
  scanf("%lld",&L);
  l=L/3.0;

  printf("%.10f\n",l*l*l);
	return 0;
}