#include<bits/stdc++.h>
#define ll long long
using namespace std;
int a[1000009];
int main(){
    int n;
    cin >> n;
    int o = 0;
    for(int i = 1; i <= n; i++){
       char c;
       cin >> c;
       a[i] = c - '1';
       if(a[i] == 1) o = 1;

    }
    if(!o){
        for(int i = 1 ;i <= n; i++){
            a[i] /= 2;
        }
    }
    int x = 0;
    for(int i = 0; i < n; i++){
        if((i & (n - 1 - i)) == 0) x ^= (a[i+1]&1);
    }
    if(!o) cout << 2*x << endl;
    else cout << x << 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<bits/stdc++.h>
#define ll long long
using namespace std;
int a[1000009];
int main(){
    int n;
    cin >> n;
    int o = 0;
    for(int i = 1; i <= n; i++){
       char c;
       cin >> c;
       a[i] = c - '1';
       if(a[i] == 1) o = 1;

    }
    if(!o){
        for(int i = 1 ;i <= n; i++){
            a[i] /= 2;
        }
    }
    int x = 0;
    for(int i = 0; i < n; i++){
        if((i & (n - 1 - i)) == 0) x ^= (a[i+1]&1);
    }
    if(!o) cout << 2*x << endl;
    else cout << x << endl;
}

```