N !

Description

Given an integer N(0 ≤ N ≤ 10000), your task is to calculate N! 

Input

One N in one line, process to the end of file. 

Output

For each N, output N! in one line. 

Sample Input

1
2
3

Sample Output

1
2
6

#include <cstring>
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <stack>
#include <vector>
#include <queue>
#include <cmath>
using namespace std;
#define N 10005
typedef long long LL;
void jiec(int a[],int num)
{
    int i;
    for(i=0;i<N;i++)
        a[i] = a[i] * num;
    
    for(i=0;i<N;i++)
    {
        a[i+1] += a[i] / 10000;
        a[i] = a[i] % 10000;
    }
}
void put(int a[])
{
    int i = N-1;

    while(!a[i]) i--;
    printf("%d",a[i]);

    for(i--;i>=0;i--)
        printf("%04d",a[i]);
        
        printf("
");
}
int main()
{
    int n;
    int a[N];
    while(scanf("%d", &n)!=EOF)
    {
        memset(a,0,sizeof(a));

        a[0] = 1;

        for(int i=1; i<=n; i++)
            jiec(a,i);

            put(a);

    }
    return 0;
}