Codeforces Round #260 (Div. 2) B. Fedya and Maths

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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:

$\text{[math]}$

数学问题，题意很简单；其实代码也很简单

我最开始做的时候真的吧n的多少次方算出来了。由于之前整理过这样的模板。感觉很爽，但是中途wa了。但发现了这个数只要是4的倍数时，答案就是4，否则就是0；

没有证明，可以简单的大表，观察就可以得出结果。这样代码就简单了

#include<iostream>
#include<fstream>
#include<iomanip>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cstdlib>
#include<cmath>
#include<set>
#include<map>
#include<queue>
#include<stack>
#include<string>
#include<vector>
#include<sstream>
#include<cassert>
using namespace std;
#ifdef __int64
typedef __int64 LL;
#else
typedef long long LL;
#endif#define nnum 1000005
#define nmax 31625int main() {int a, c;char ch[nnum];while (~scanf("%s",ch)) {int len=strlen(ch);int a=ch[len-1]-'0';int b=ch[len-2]-'0';int sum=a+b*10;if(sum%4==0){cout<<"4"<<endl;}elsecout<<"0"<<endl;}return 0;
}

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