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[+];//注意拿数组进行储存
int main()
{
    scanf("%s",str);
    int len = strlen(str);
    int sum = (str[len-]-'')*+str[len-]-'';
    if(sum%==)puts("");//做题前先打表
    else puts("");
    return ;
}