#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int N=2.2e6;
char c[N];
int a[N],n,b[N];
int C(int n,int k)
{
	if((n&k)==k)return 1;return 0;
 } 
int main()
{//cout<<C(5,0)<<endl;
	cin>>n;scanf("%s",c+1);
	for(int i=1;i<=n;i++)a[i]=c[i]-'0';int fg=0;n--;
	for(int i=1;i<=n;i++){
		b[i]=abs(a[i]-a[i+1]);
		if(b[i]==1)fg=1;
	}int ans=0;
	if(fg)
	{
		for(int i=1;i<=n;i++)b[i]=b[i]%2;
		for(int i=1;i<=n;i++)ans=ans^(C(n-1,i-1)*b[i]);
		cout<<ans;
		return 0;
	}
	for(int i=1;i<=n;i++)b[i]=b[i]/2;
	//for(int i=1;i<=n;i++)cout<<b[i]<<" ";puts("");
	for(int i=1;i<=n;i++)ans=ans^(C(n-1,i-1)*b[i]);
	cout<<ans*2;
}

### Prompt

Please create a solution in cpp 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;
#define ll long long
const int N=2.2e6;
char c[N];
int a[N],n,b[N];
int C(int n,int k)
{
	if((n&k)==k)return 1;return 0;
 } 
int main()
{//cout<<C(5,0)<<endl;
	cin>>n;scanf("%s",c+1);
	for(int i=1;i<=n;i++)a[i]=c[i]-'0';int fg=0;n--;
	for(int i=1;i<=n;i++){
		b[i]=abs(a[i]-a[i+1]);
		if(b[i]==1)fg=1;
	}int ans=0;
	if(fg)
	{
		for(int i=1;i<=n;i++)b[i]=b[i]%2;
		for(int i=1;i<=n;i++)ans=ans^(C(n-1,i-1)*b[i]);
		cout<<ans;
		return 0;
	}
	for(int i=1;i<=n;i++)b[i]=b[i]/2;
	//for(int i=1;i<=n;i++)cout<<b[i]<<" ";puts("");
	for(int i=1;i<=n;i++)ans=ans^(C(n-1,i-1)*b[i]);
	cout<<ans*2;
}

```