#include <bits/stdc++.h>
using namespace std;

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    int N; string S; cin >> N >> S;
    vector<int> a(N);
    int one=0,ans=0;
    for (int i=0;i<N;++i){
        a[i]=S[i]-'1';
        one|=(a[i]==1);
    }
    for (int i=0;i<N;++i) if (((N-1)&i)==i) ans^=(one?a[i]%2:a[i]/2);
    cout << ans*(one?1:2) << '\n';
}
### Prompt

Generate 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;

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    int N; string S; cin >> N >> S;
    vector<int> a(N);
    int one=0,ans=0;
    for (int i=0;i<N;++i){
        a[i]=S[i]-'1';
        one|=(a[i]==1);
    }
    for (int i=0;i<N;++i) if (((N-1)&i)==i) ans^=(one?a[i]%2:a[i]/2);
    cout << ans*(one?1:2) << '\n';
}
```