最后的 0 是因为 5 和 2 产生的对吧，而且 2 的因子还比 5 多，因此，我们只需要知道 5 的因子有多少个就行了
那么我们来看 5 的因子有多少个
以 58 ! 为例：
58 ! = 1 ⋅ 2 ⋅ 3 ⋅ 4 ⋅ 5 ⋅ 6 ⋅ 7 ⋅ 8 ⋅ 9 ⋅ 10 ⋅ 11 ⋅ 12 ⋅ 13 ⋅ 14 ⋅ 15 ⋅ 16 ⋅ 17 ⋅ 18 ⋅ 19 ⋅ 20 ⋅ 21 ⋅ 22 ⋅ 23 ⋅ 24 ⋅ 25 26 ⋅ 27 ⋅ 28 ⋅ 29 ⋅ 30 ⋅ 31 ⋅ 32 ⋅ 33 ⋅ 34 ⋅ 35 ⋅ 36 ⋅ 37 ⋅ 38 ⋅ 39 ⋅ 40 ⋅ 41 ⋅ 42 ⋅ 43 ⋅ 44 ⋅ 45 ⋅ 46 ⋅ 47 ⋅ 48 ⋅ 49 ⋅ 50 ⋅ 51 ⋅ 52 ⋅ 53 ⋅ 54 ⋅ 55 ⋅ 56 ⋅ 57 ⋅ 58
标红的都是有因子 5 的，而且是每 5 个就是一个含有 5 的因子，假如 5 的因子有 k 个，那么把这些个因子 5 都提取出来， k = 58 5 = 11 个
于是数列变成了

54 ! = 1 ⋅ 2 ⋅ 3 ⋅ 4 ⋅ 1 ⋅ 6 ⋅ 7 ⋅ 8 ⋅ 9 ⋅ 2 ⋅ 11 ⋅ 12 ⋅ 13 ⋅ 14 ⋅ 3 ⋅ 16 ⋅ 17 ⋅ 18 ⋅ 19 ⋅ 4 ⋅ 21 ⋅ 22 ⋅ 23 ⋅ 24 ⋅ 5 26 ⋅ 27 ⋅ 28 ⋅ 29 ⋅ 6 ⋅ 31 ⋅ 32 ⋅ 33 ⋅ 34 ⋅ 7 ⋅ 36 ⋅ 37 ⋅ 38 ⋅ 39 ⋅ 8 ⋅ 41 ⋅ 42 ⋅ 43 ⋅ 44 ⋅ 9 ⋅ 46 ⋅ 47 ⋅ 48 ⋅ 49 ⋅ 10 ⋅ 51 ⋅ 52 ⋅ 53 ⋅ 54 ⋅ 11 ⋅ 56 ⋅ 57 ⋅ 58

蓝色的就是提取了因子 5 的，红色就是提取之后还有因子 5 的
但是此时的间隔就不再是 5 了，而是 5 × 5 = 25 了
所以 k = 11 + 58 25 = 13 个
提取之后的数列是这样的：
54 ! = 1 ⋅ 2 ⋅ 3 ⋅ 4 ⋅ 1 ⋅ 6 ⋅ 7 ⋅ 8 ⋅ 9 ⋅ 2 ⋅ 11 ⋅ 12 ⋅ 13 ⋅ 14 ⋅ 3 ⋅ 16 ⋅ 17 ⋅ 18 ⋅ 19 ⋅ 4 ⋅ 21 ⋅ 22 ⋅ 23 ⋅ 24 ⋅ 1 26 ⋅ 27 ⋅ 28 ⋅ 29 ⋅ 6 ⋅ 31 ⋅ 32 ⋅ 33 ⋅ 34 ⋅ 7 ⋅ 36 ⋅ 37 ⋅ 38 ⋅ 39 ⋅ 8 ⋅ 41 ⋅ 42 ⋅ 43 ⋅ 44 ⋅ 9 ⋅ 46 ⋅ 47 ⋅ 48 ⋅ 49 ⋅ 2 ⋅ 51 ⋅ 52 ⋅ 53 ⋅ 54 ⋅ 11 ⋅ 56 ⋅ 57 ⋅ 58

好现在就没有因子 5 了
看这个 k 怎么来的？后一次比前一次要多除 5
那么其实阔以看成每次的结果来除以 5
也就是：
58 5 = 11
k + = 11

11 5 = 2
k + = 2

2 5 = 0
好，到此结束 k = 12
其实这个就是 n ! 求某个质因子的个数的方法，一直除，除的结果加到个数里面
那么我们最后这个数列乘出来最后就没有 0 了对吧
那么这个数列对 5 取余是不等于 0 的吧~
那么我们就让其对 5 取余，并且每次不看蓝色的数
这个时候就有规律了：
( 1 ⋅ 2 ⋅ 3 ⋅ 4 ) % 5 = 4
( 6 ⋅ 7 ⋅ 8 ⋅ 4 ) % 9 = ( 1 ⋅ 2 ⋅ 3 ⋅ 4 ) % 5 = 4
发现什么木有，有循环，而且都是 4
这种循环有多少个呢？

第一次除以 5 的时候有 10 个对吧，而且还有 ( 51 ⋅ 52 ⋅ 53 ) 这3个数不属于循环里的，而这三个数的乘积 想当于 3 ! ，而 53 % 5 = 3

第二次有 2 个，而且有一个 55 不属于第二次循环里面的，相当于 1 !
而 11 % 5 = 1

所以对 5 取余就是：
( 1 ⋅ 2 ⋅ 3 ⋅ 4 ) k ∗ 每 次 的 余 数 的 阶 乘
结论就是：（在对 5 取余的情况下）

2 k ⋅ n ! k 5 ≡ ( 1 ⋅ 2 ⋅ 3 ⋅ 4 ) k ∗ 每 次 的 余 数 的 阶 乘

而 ( 1 ⋅ 2 ⋅ 3 % 5 = 1 ) 所以这一坨就阔以不要了
结论变成：
2 k ⋅ n ! k 5 ≡ ( 4 ) k ∗ 每 次 的 余 数 的 阶 乘

然后再与左边的 2 k 约一哈：
n ! k 5 ≡ ( 2 ) k ∗ 每 次 的 余 数 的 阶 乘

而 2 4 % 5 = 1
所以还阔以简化一哈。。。
结论变成：
n ! k 5 ≡ ( 2 ) k % 4 ∗ 每 次 的 余 数 的 阶 乘

因此， n ! 把因子 5 去掉后对 5 取余就知道是多少了，假设是 x
而其中 2 的因子比 5 多，那么里面肯定还有 2 的因子，所以对 2 取余为 0
所以得到两个同余方程：
n ! 5 k ≡ x （mod 5）
n ! 5 k ≡ 0 （mod 2）
那么就能得到对 10 取余是多少了，也就是答案
r e s ≡ n ! 10 5 （mod 10） 
/* 阶乘最后一位非零数 */
#include"bits/stdc++.h"
using namespace std;
int exgcd(int a, int b, int &x, int &y,int c)
{
    if (b == 0)
    {
        x = c;
        y = 0;
        return a;
    }

    int d = exgcd(b, a % b, x, y,c);
    int t = x;
    x = y;
    y = t - a / b * y;
    return d;
}
int main()
{
    long long fac[10]={1,1,2,6,24};
    long long N;
    while(cin>>N)
    {
        int k=0;
        long long ans=1;
        while(N)
        {
            k+=N/5;
            ans*=fac[N%5];
            ans%=5;
            N/=5;
        }
        k%=4;
        ans*=1<<k;
        ans%=5;
        int x,y;
        exgcd(5,2,x,y,-ans);
        int res=x*5+ans;
        res=(res%10+10)%10;
        cout<<res<<endl;
    }
}