CF#260 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:

不懂什麼費馬小定理。打表之後發現：n%4=0的時候輸出4，
其餘輸出0，但n資料太大了.我們知道：100以上的數可以分解成100*x+y者只需考慮y就可以了，即只需考慮最後兩位。這就好辦了。讀入字串得到最後兩位模數就可以了

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<limits.h>#include<cmath>using namespace std;const int maxn=1e6;char s[maxn];int main(){    while(~scanf("%s",s))    {        int len=strlen(s);        int sum=0;        sum+=s[len-1]-'0'+(s[len-2]-'0')*10;        if(sum%4==0)            cout<<4<<endl;        else            cout<<0<<endl;    }    return 0;}