#include<bits/stdc++.h>
using namespace std;
using LL = long long;
int C(int M, int N){
    for(; N; N >>= 1, M >>= 1) if((N & 1) > (M & 1)) return 0;
    return 1;
}
int main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    int N, ans = 0, k = 1;
    cin >> N;
    string s;
    cin >> s;
    for(int i = 1; i < N; i += 1) s[i - 1] = abs(s[i] - s[i - 1]);
    s.pop_back();
    k += not count(s.begin(), s.end(), 1);
    for(int i = 0; i < N - 1; i += 1) ans += C(N - 2, i) * s[i] / k;
    cout << ans % 2 * k;
    return 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;
using LL = long long;
int C(int M, int N){
    for(; N; N >>= 1, M >>= 1) if((N & 1) > (M & 1)) return 0;
    return 1;
}
int main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    int N, ans = 0, k = 1;
    cin >> N;
    string s;
    cin >> s;
    for(int i = 1; i < N; i += 1) s[i - 1] = abs(s[i] - s[i - 1]);
    s.pop_back();
    k += not count(s.begin(), s.end(), 1);
    for(int i = 0; i < N - 1; i += 1) ans += C(N - 2, i) * s[i] / k;
    cout << ans % 2 * k;
    return 0;
}

```