Fedya and Maths

B. Fedya and Maths

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

 

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:

这道题上来直接打表找规律

发现能整除4的 n 算出来是4，其他的n为0。要注意输入很大，要用字符串。

这里处理就出多说了，假设一个数字n前面n-2个数为 x 后面两个数字为 y ，则这个数 n=x*100+y

前面必然能除4，判断后面两位能否整除四即可

#include<cstdio>
#include<cmath>
#include<cstring>
#include<stdlib.h>
using namespace std;
int main()
{
    char str[100005],a,b;
    int ans;
    scanf("%s",str);
    int len=strlen(str);
    if(len>=2)
    {
        a=str[len-1];
        b=str[len-2];
        ans=(b-'0')*10+a-'0';
    }
    else
        ans=str[len-1]-'0';

    if(ans%4==0)
        printf("4
");
    else
        printf("0
");
    return 0;
}