#include<cstdio>
#define N 1000001
inline int abs(int x){return x<0?-x:x;}
int b[3],i,n,x=0;char a[N];
int main()
{
	scanf("%d%s",&n,a);
	for(i=0,n--;i<n;i++)b[a[i]=abs(a[i]-a[i+1])%3]=1;
	for(i=0,n--;i<=n;i++)if((i&n)==i)x^=a[i];
	return 0&printf("%d\n",x&~(b[1]<<1));
}
### Prompt

Develop a solution in CPP to the problem described below:

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

### Response

```cpp
#include<cstdio>
#define N 1000001
inline int abs(int x){return x<0?-x:x;}
int b[3],i,n,x=0;char a[N];
int main()
{
	scanf("%d%s",&n,a);
	for(i=0,n--;i<n;i++)b[a[i]=abs(a[i]-a[i+1])%3]=1;
	for(i=0,n--;i<=n;i++)if((i&n)==i)x^=a[i];
	return 0&printf("%d\n",x&~(b[1]<<1));
}
```