Codeforces Round #260 (Div. 2) B

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.

Examples

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:

题意：看题目中的公式

解法：打表找规律

?

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35

#include<bits/stdc++.h> using namespace std; class P {     public:     int n,m; }; bool cmd(P x,P y) {     return x.n<y.n; } int main() {     long long ans=0;     string s;     cin>>s;     if(s.length()==1)     {      ans+=s[s.length()-1]-'0';     }     else     {         ans+=((s[s.length()-2]-'0')*10+s[s.length()-1]-'0');     }   //  cout<<ans<<endl;     if(ans%4)     {         cout<<"0"<<endl;     }     else     {         cout<<"4"<<endl;     }     return 0; }