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

#define LL long long

int main() {
    ios::sync_with_stdio(0), cin.tie(0);
    
    int n, cur = 0; cin >> n; string s; cin >> s;
    int a[n-1], cnt[3]={};
    for(int i = 0; i < n-1; i++) {
        a[i] = abs(s[i+1] - s[i]);
        cnt[a[i]]++;
    }

    int mul = 1, res = 0;
    if(cnt[1] == 0) {
        mul++;
        for(int &x : a) x >>= 1;
    }

    for(int i = 0; i < n-1; i++) if(((n-2) & i) == i) res ^= a[i];
    cout << (res & 1) * mul; 
}
### 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;

#define LL long long

int main() {
    ios::sync_with_stdio(0), cin.tie(0);
    
    int n, cur = 0; cin >> n; string s; cin >> s;
    int a[n-1], cnt[3]={};
    for(int i = 0; i < n-1; i++) {
        a[i] = abs(s[i+1] - s[i]);
        cnt[a[i]]++;
    }

    int mul = 1, res = 0;
    if(cnt[1] == 0) {
        mul++;
        for(int &x : a) x >>= 1;
    }

    for(int i = 0; i < n-1; i++) if(((n-2) & i) == i) res ^= a[i];
    cout << (res & 1) * mul; 
}
```