#include<iostream>
#include<cstdio>
using namespace std;
int n,a[1000100],b[1000100];
int aabs(int x){ return x<0?-x:x;}
int calc(int x,int y,int dp)
{
	if(y>x) return 0;
	if(dp==0) return 1;
	int ln=1<<dp;
	if(x>ln) x-=ln;
	if(y>ln) y-=ln;
	return calc(x,y,dp-1);
}
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
	scanf("%1d",&a[i]);
	if(n==1) return  printf("%d\n",a[1]),0;
	for(int i=1;i<n;i++)
	b[i]=aabs(a[i]-a[i+1]);
	int fl=1;
	for(int i=1;i<n;i++)
	if(b[i]==1) fl=0;
	int ans=0;
	if(!fl)
	{
		for(int i=1;i<n;i++)
		if(b[i]==1) ans^=calc(n-1,i,20);
		printf("%d\n",ans);
	}
	else
	{
		for(int i=1;i<n;i++)
		if(b[i]==2) ans^=calc(n-1,i,20);
		printf("%d\n",ans*2);
	}
	return 0;
}
### Prompt

Your challenge is to write a cpp solution to the following problem:

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<iostream>
#include<cstdio>
using namespace std;
int n,a[1000100],b[1000100];
int aabs(int x){ return x<0?-x:x;}
int calc(int x,int y,int dp)
{
	if(y>x) return 0;
	if(dp==0) return 1;
	int ln=1<<dp;
	if(x>ln) x-=ln;
	if(y>ln) y-=ln;
	return calc(x,y,dp-1);
}
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
	scanf("%1d",&a[i]);
	if(n==1) return  printf("%d\n",a[1]),0;
	for(int i=1;i<n;i++)
	b[i]=aabs(a[i]-a[i+1]);
	int fl=1;
	for(int i=1;i<n;i++)
	if(b[i]==1) fl=0;
	int ans=0;
	if(!fl)
	{
		for(int i=1;i<n;i++)
		if(b[i]==1) ans^=calc(n-1,i,20);
		printf("%d\n",ans);
	}
	else
	{
		for(int i=1;i<n;i++)
		if(b[i]==2) ans^=calc(n-1,i,20);
		printf("%d\n",ans*2);
	}
	return 0;
}
```