Codeforces 456B Fedya and Maths 打表找规律

Description

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 Input

Input

4

Output

4

Input

124356983594583453458888889

Output

0

Hint

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

Note to the first sample:

题意：  给你N  让你求 (1ⁿ + 2ⁿ + 3ⁿ + 4ⁿ) mod 5

题解：  打个表 发现了规律

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <queue>
#include<vector>
#include <map>
using namespace std ;
typedef long long ll;

const int N=100+10;
const int maxn = 1000000;

char s[maxn];
int main()
{
    scanf("%s",s);
    int len=strlen(s);
    int a=s[len-1]-'0';
    int b=s[len-2]-'0';
    int c=a+b*10;
    if(c%4==0)
        printf("4
");
    else
        printf("0
");
    return 0;
}