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

#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;
}