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条从北向南的路和n条从东向西的路，共n^2个交点，共5把椅子，放在交点，然后每条路上最多只能有一把椅子

求放椅子的方法数 方法比较笨，每次循环放一把椅子

#include <iostream>#include<cstdio>#include<cstring>#include<cmath>#define maxn 110using 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;}