Fedya and Maths

题目：

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 1

Inputcopy	Outputcopy
4	4

Sample 2

Inputcopy	Outputcopy
124356983594583453458888889	0

Note

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

Note to the first sample:

思路：

用一个数组来存储n的每一位，高精度求出n模4的结果，根据结果进行判断。

代码：

#include<string.h>
#include<stdio.h>
char a[100005] = { 0 };
int main()
{
	//输入数字n 
	scanf("%s",a);
	int len = strlen(a);
	for (int i = 0; i < len; i++) a[i] = a[i] - '0';
	int up = 0,x = 0, tmp = 0, ans = 0;
	//计算n模4的结果 
	for (int i = 0; i < len; i++)
	{
		tmp = a[i] + up * 10;
		up = tmp % 4;
	}
	x = up;
	//根据n模4的结果做出判断 
	if(x == 0) ans = 4;
	printf("%d",ans);
	return 0;
 }