GDUT - 专题学习4 B - Fedya and Maths

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 ≤ 10^(10^5)). The number doesn't contain any leading zeroes.

Output

Print the value of the expression without leading zeros.

Sample 1

4

4

Sample 2

124356983594583453458888889

0

Note

Operation x mod y means taking remainder after division x by y.

Note to the first sample:

一开始毫无思路，准备打表试一下，发现整除4的数结果都是4，其余的数都是0，但题目的数特别大，longlong都存不下，其实用string存就可以了，然后判断最后两位是否整除4即可，最后特判一下0、4、8。

#include
#include
using namespace std;
string s;
int main()
{
	cin >> s;
	int t = s.length();
	if (t == 1)
	{
		if ((s[0] - '0') % 4 == 0)
		{
			cout << 4;
		}
		else
		{
			cout << 0;
		}
	}
	else
	{
		if ((s[t - 1] - '0' + (s[t - 2] - '0') * 10) % 4 == 0)
		{
			cout << 4;
		}
		else
		{
			cout << 0;
		}
	}
	return 0;
}