#include<bits/stdc++.h>
using namespace std;
char s[1000010];

inline int C(int n, int m) { return (n & m) == m; }

int main() {
    int n;
    scanf("%d%s", &n, s + 1);
    if(n == 1) putchar(s[1]), puts("");
    else {
        --n;
        int find1 = 0;
        for(int j = 1; j <= n; j++)
            s[j] = abs(s[j] - s[j + 1]), find1 |= s[j] == 1;
        if(!find1) for(int i = 1; i <= n; i++) s[i] >>= 1;
        find1 ^= 1;
        int t = 0;
        for(int j = 1; j <= n; j++) t ^= C(n - 1, j - 1) * (s[j] & 1);
        cout << (int(t) << find1) << endl;
    }
    return 0;
}
### Prompt

Construct a cpp code solution to the problem outlined:

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;
char s[1000010];

inline int C(int n, int m) { return (n & m) == m; }

int main() {
    int n;
    scanf("%d%s", &n, s + 1);
    if(n == 1) putchar(s[1]), puts("");
    else {
        --n;
        int find1 = 0;
        for(int j = 1; j <= n; j++)
            s[j] = abs(s[j] - s[j + 1]), find1 |= s[j] == 1;
        if(!find1) for(int i = 1; i <= n; i++) s[i] >>= 1;
        find1 ^= 1;
        int t = 0;
        for(int j = 1; j <= n; j++) t ^= C(n - 1, j - 1) * (s[j] & 1);
        cout << (int(t) << find1) << endl;
    }
    return 0;
}
```