#include <bits/stdc++.h>
#define rep(i,n)for(int i=0;i<(n);i++)
using namespace std;
typedef long long ll;

int a[2000000];
int b[2000000];
int main(){
	int n;cin>>n;
	rep(i,n)scanf("%1d",&a[i]);
	set<int>se;
	rep(i,n-1){
		b[i]=abs(a[i]-a[i+1]);
		se.insert(b[i]);
	}
	n--;
	bool flag=false;
	if(se.count(1)){
		rep(i,n)b[i]%=2;
	}
	else{
		flag=true;
		rep(i,n)b[i]/=2;
	}
	int ans=0;
	rep(i,n){
		if(n-1==(i|(n-1-i)))ans^=b[i];
	}
	if(flag)ans*=2;
	cout<<ans<<endl;
}
### Prompt

Create a solution in Cpp for 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>
#define rep(i,n)for(int i=0;i<(n);i++)
using namespace std;
typedef long long ll;

int a[2000000];
int b[2000000];
int main(){
	int n;cin>>n;
	rep(i,n)scanf("%1d",&a[i]);
	set<int>se;
	rep(i,n-1){
		b[i]=abs(a[i]-a[i+1]);
		se.insert(b[i]);
	}
	n--;
	bool flag=false;
	if(se.count(1)){
		rep(i,n)b[i]%=2;
	}
	else{
		flag=true;
		rep(i,n)b[i]/=2;
	}
	int ans=0;
	rep(i,n){
		if(n-1==(i|(n-1-i)))ans^=b[i];
	}
	if(flag)ans*=2;
	cout<<ans<<endl;
}
```