Codeforces Round #260 (Div. 2) B. 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:

#include <cstdio>

#include <cstring>

using namespace std;

char str[100000+10];//注意拿数组进行储存

int main()

{

    scanf("%s",str);

    int len = strlen(str);

    int sum = (str[len-2]-'0')*10+str[len-1]-'0';

    if(sum%4==0)puts("4");//做题前先打表

    else puts("0");

    return 0;

}