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<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
using namespace std;
int n;
char s[1000010];
int num(int x)
{
	int res=0;
	while(x%2==0)res++,x/=2;
	return res;
}
int a[1000010];
int main()
{
	int n;scanf("%d",&n);
	scanf("%s",s);
	n--;
	for(int i=0;i<n;i++)
		a[i]=max(s[i]-s[i+1],s[i+1]-s[i]);
	int cur=0;
	bool flag=false;
	int res=0;
	for(int i=0;i<n;i++)
	{
		if(i)cur+=num(n-i)-num(i);
		if(a[i]==1)
			res^=(cur==0),flag=true;
	}
	if(flag)printf("%d\n",res);
	else{
		cur=0;res=0;
		for(int i=0;i<n;i++)
		{
			if(i)cur+=num(n-i)-num(i);
			if(a[i]==2)
				res^=(cur==0);
		}
		printf("%d\n",res*2);
	}
	return 0;
}