Codeforces 630H Benches

H. Benches time limit per test 0.5 seconds memory limit per test 64 megabytes input standard input output standard output

The city park of IT City contains n east to west paths and n north to south paths. Each east to west path crosses each north to south path, so there are n2 intersections.

The city funded purchase of five benches. To make it seems that there are many benches it was decided to place them on as many paths as possible. Obviously this requirement is satisfied by the following scheme: each bench is placed on a cross of paths and each path contains not more than one bench.

Help the park administration count the number of ways to place the benches.

Input

The only line of the input contains one integer n (5 ≤ n ≤ 100) — the number of east to west paths and north to south paths.

Output

Output one integer — the number of ways to place the benches.

Examples input

5
      

output

120      

n條從北向南的路和n條從東向西的路，共n^2個交點，共5把椅子，放在交點，然後每條路上最多隻能有一把椅子

求放椅子的方法數 方法比較笨，每次循環放一把椅子

#include <iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#define maxn 110
using namespace std;
int main()
{
    int n;
    long long sum;
    while(~scanf("%d",&n))
    {
        sum=1;
        int flag1=0,flag2=0,flag3=0,flag4=0;
        for(int i=0;i<5;++i)
        {
            sum*=(n-i)*(n-i);
            if(!flag1&&sum%2==0)
            {
                flag1=1;
                sum=sum/2;
            }
            if(!flag2&&sum%3==0)
            {
                flag2=1;
                sum=sum/3;
            }
            if(!flag3&&sum%4==0)
            {
                flag3=1;
                sum=sum/4;
            }
            if(!flag4&&sum%5==0)
            {
                flag4=1;
                sum=sum/5;
            }
        }
        printf("%I64d\n",sum);
    }
    return 0;
}