Given is a sequence of N digits a_1a_2\ldots a_N, where each element is 1, 2, or 3. Let x_{i,j} defined as follows:

* x_{1,j} := a_j \quad (1 \leq j \leq N)
* x_{i,j} := | x_{i-1,j} - x_{i-1,j+1} | \quad (2 \leq i \leq N and 1 \leq j \leq N+1-i)



Find x_{N,1}.

Constraints

* 2 \leq N \leq 10^6
* a_i = 1,2,3 (1 \leq i \leq N)

Input

Input is given from Standard Input in the following format:


N
a_1a_2\ldotsa_N


Output

Print x_{N,1}.

Examples

Input

4
1231


Output

1


Input

10
2311312312


Output

0
#include<bits/stdc++.h>
#define INF 0x3f3f3f3f
const int MOD=1e9+7;
const double eps=1e-8;
typedef long long ll;
//orz yht
using namespace std;
int n,a[1000005],p,vis[5];
char s[1000005];
int main(){
//	freopen(".in","r",stdin);
//	freopen(".out","w",stdout);
	cin>>n;
	scanf("%s",s+1);
	for(int i=1;i<n;i++){
		a[i]=abs(s[i]-s[i+1]);
		vis[a[i]]=1;
	}
	if(!vis[1]){
		for(int i=1;i<n;i++)
		a[i]/=2;
	}
	else{
		for(int i=1;i<n;i++)
		a[i]%=2;
	}
	int nn=n-2;
	for(int i=0;i<n-1;i++){
		if((nn&i)==i)
		p+=a[i+1];
	}
	int ans=p%2;
	if(!vis[1]){
		if(ans)
		ans=2;
	}
	cout<<ans;
	return 0;
}