#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

int n;
int arr[1000002];
int ans, exist1;
int isOdd;

int main(){
    scanf("%d", &n);
    for(int i=1; i<=n; i++){
        scanf("%1d", &arr[i]);
        arr[i]--;

        if(arr[i] == 1) exist1 = 1, isOdd ^= !((i-1) & (n-i));
    }
    if(isOdd){ printf("1"); return 0; }

    if(!exist1){
        for(int i=1; i<=n; i++){
            if(arr[i] == 2) isOdd ^= !((i-1) & (n-i));
        }
        if(isOdd) printf("2");
        else printf("0");
        return 0;
    }

    printf("0");
}

### 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>

using namespace std;

typedef long long ll;

int n;
int arr[1000002];
int ans, exist1;
int isOdd;

int main(){
    scanf("%d", &n);
    for(int i=1; i<=n; i++){
        scanf("%1d", &arr[i]);
        arr[i]--;

        if(arr[i] == 1) exist1 = 1, isOdd ^= !((i-1) & (n-i));
    }
    if(isOdd){ printf("1"); return 0; }

    if(!exist1){
        for(int i=1; i<=n; i++){
            if(arr[i] == 2) isOdd ^= !((i-1) & (n-i));
        }
        if(isOdd) printf("2");
        else printf("0");
        return 0;
    }

    printf("0");
}

```