CF 456B(Fedya and Maths-找循环节)

B. Fedya and Maths

time limit per test

memory limit per test

input

output

Fedya studies in a gymnasium. Fedya's maths hometask is to calculate the following expression:

(1n + 2n + 3n + 4n) mod 5

n. Fedya managed to complete the task. Can you? Note that given number n

Input

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

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

Note to the first sample:

易发现本题的循环节 [0,0,0,4],[0,0,0,4],...

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Lson (x<<1)
#define Rson ((x<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define MAXN (100000+10)
long long mul(long long a,long long b){return (a*b)%F;}
long long add(long long a,long long b){return (a+b)%F;}
long long sub(long long a,long long b){return (a-b+(a-b)/F*F+F)%F;}
typedef long long ll;
char c[MAXN];
int main()
{
//  freopen("b.in","r",stdin);
//  freopen(".out","w",stdout);
  scanf("%s",c+1);
  int n=strlen(c+1);
  int t;
  if (n==1) t=c[1]-'0';
  else t=c[n]-'0'+(c[n-1]-'0')*10;
  
  if (t%4==0) cout<<"4\n";
  else cout<<"0\n";
  
  
  return 0;
}