CF#260 B. Fedya and Maths

Fedya studies in a gymnasium. Fedya‘s maths hometask is to calculate the following expression:

(1n?+?2n?+?3n?+?4n) mod 5

for given value of n. Fedya managed to complete the task. Can you? Note that given number n can be extremely large (e.g. it can exceed any integer type of your programming language).

Input

The single line contains a single integer n (0?≤?n?≤?10105). The number doesn‘t contain any leading zeroes.

Output

Print the value of the expression without leading zeros.

Sample test(s)

input
4

output
4

input
124356983594583453458888889

output
0

Note

Operation x mod y means taking remainder after division x by y.

Note to the first sample:

$bubuko.com,布布扣$

不懂什么费马小定理。打表之后发现：n%4=0的时候输出4，
其余输出0，但n数据太大了.我们知道：100以上的数可以分解成100*x+y者只需考虑y就可以了，

即只需考虑最后两位。这就好办了。读入字符串得到最后两位取模就可以了

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<limits.h>
#include<cmath>
using namespace std;
const int maxn=1e6;
char s[maxn];
int main()
{
    while(~scanf("%s",s))
    {
        int len=strlen(s);
        int sum=0;
        sum+=s[len-1]-'0'+(s[len-2]-'0')*10;
        if(sum%4==0)
            cout<<4<<endl;
        else
            cout<<0<<endl;
    }
    return 0;
}