#include <iostream>
#include <string>

using namespace std;

int main(){
    int N;
    string a;
    cin >> N >> a;
    auto isOdd = [&](){
        int res = 0;
        for(int i=0;i<N;i++){
            if(a[i] == '2' && ((N-1)&i) == i) res = 1-res;
        }
        return res;
    };
    if(a.find('2') != string::npos){
        cout << (isOdd() ? 1 : 0) << endl;
    } else {
        for(auto& c : a) if(c == '3') c = '2';
        cout << (isOdd() ? 2 : 0) << endl;
    }
}
### Prompt

Develop a solution in CPP to the problem described below:

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 <string>

using namespace std;

int main(){
    int N;
    string a;
    cin >> N >> a;
    auto isOdd = [&](){
        int res = 0;
        for(int i=0;i<N;i++){
            if(a[i] == '2' && ((N-1)&i) == i) res = 1-res;
        }
        return res;
    };
    if(a.find('2') != string::npos){
        cout << (isOdd() ? 1 : 0) << endl;
    } else {
        for(auto& c : a) if(c == '3') c = '2';
        cout << (isOdd() ? 2 : 0) << endl;
    }
}
```