#include <bits/stdc++.h>
using namespace std;
const int N=1000100;
int f[N],g[N],h[N],a[N],n,x,b[N];
char s[N];
int main()
{
	scanf("%d",&n);
	for (int i=1;i<=1000000;i++)
		if (i%2==0) f[i]=f[i/2]+1;
	for (int i=1;i<n;i++)
	{
		g[i]=g[i-1]+f[n-i]-f[i];
		if (g[i]) h[i]=0; else h[i]=1;
	}
	h[0]=1;
	scanf("%s",s+1);
	for (int i=1;i<=n;i++)
	{
		a[i]=s[i]-'1';
		b[a[i]]++;
	}
	//for (int i=0;i<n;i++) printf("%d ",h[i]);
	if (b[1]==0)
	{
		for (int i=1;i<=n;i++)
			if (a[i]==2) x+=h[i-1];
		if (x%2==0) printf("%d",0); else printf("%d",2);
	}
	else
	{
		for (int i=1;i<=n;i++)
			if (a[i]==1) x+=h[i-1];
		if (x%2==0) printf("%d",0); else printf("%d",1);
	}
} 
### 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 <bits/stdc++.h>
using namespace std;
const int N=1000100;
int f[N],g[N],h[N],a[N],n,x,b[N];
char s[N];
int main()
{
	scanf("%d",&n);
	for (int i=1;i<=1000000;i++)
		if (i%2==0) f[i]=f[i/2]+1;
	for (int i=1;i<n;i++)
	{
		g[i]=g[i-1]+f[n-i]-f[i];
		if (g[i]) h[i]=0; else h[i]=1;
	}
	h[0]=1;
	scanf("%s",s+1);
	for (int i=1;i<=n;i++)
	{
		a[i]=s[i]-'1';
		b[a[i]]++;
	}
	//for (int i=0;i<n;i++) printf("%d ",h[i]);
	if (b[1]==0)
	{
		for (int i=1;i<=n;i++)
			if (a[i]==2) x+=h[i-1];
		if (x%2==0) printf("%d",0); else printf("%d",2);
	}
	else
	{
		for (int i=1;i<=n;i++)
			if (a[i]==1) x+=h[i-1];
		if (x%2==0) printf("%d",0); else printf("%d",1);
	}
} 
```