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

B. Fedya and Maths

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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:

让求出 （1^n + 2^n + 3^n + 4^n ）mod 5 的结果，而n最大可以有10^5位数，只能用字符串来存，将取模分开得到

1^n mod 5 = 1 ;

2^n mod 5 = 随着n从0开始，结果为 1 2 4 3 1 2 4 3 1 2 4 3 。。。。以1 2 4 3 循环

3^n mod 5 = 随着n从0开始，结果为 1 3 4 2 1 3 4 2 1 3 4 2.。。。。以1 3 4 2 循环

4^n mod 5 = 。。。。结果为 1 4 1 4 1 4 以1 4 循环

求给出的n是奇数还是偶数，方法判断最后一位奇数偶数。再求解n mod 4 ，方法求出最后两位的对4取模的值

#include <cstdio>
#include <cstring>
char str[110000] ;
int main()
{
    int ans = 1 , l , i , k ;
    scanf("%s", str);
    l = strlen(str);
    if( (str[l-1]-'0')%2 )
        ans += 4 ;
    else
        ans += 1 ;
    if( l-1 != -1 )
        k = ( str[l-1]-'0' );
    if( l-2 != -1 )
        k = k + (str[l-2]-'0')*10 ;
    if( k%4 == 0 )
        ans += 2 ;
    else if( k%4==1 )
        ans += 5 ;
    else if( k%4 ==2 )
        ans += 8 ;
    else
        ans += 5 ;
    printf("%d\n", ans%5);
    return 0;
}