Codeforces Round #260 (Div. 2) B. Fedya and Maths(循环结)

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 numbern 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 divisionx by y.

Note to the first sample:

又是个找循环结。

数特别大并且，不难的情况应想找规律。

#include<iostream>
#include<cstring>
#include<cstdio>
#include<string>
#include<cmath>
#include<algorithm>
#define LL  __int64
#define inf 0x3f3f3f3f
using namespace std;
int main()
{
   int n,m,i,j;int k,x,y;
    while(~scanf("%I64d",&n))
    {
        n%=4;
        if(n==0)
        printf("4\n");
        else
        printf("0\n");
    }
    return 0;
}