#include <cmath>
#include <iostream>
#include <vector>

using namespace std;
int main() {
  int n;
  cin >> n;
  string s;
  cin >> s;
  vector<int> b(n - 1);
  bool has_one = false;
  for (int i = 0; i < n - 1; ++i) {
    b[i] = abs(s[i + 1] - s[i]);
    if (b[i] == 1)
      has_one = true;
  }
  int factor = 1;
  if (!has_one) {
    factor = 2;
    for (int i = 0; i < n - 1; ++i)
      b[i] >>= 1;
  }
  int sum = 0;
  for (int i = 0; i < n - 1; ++i) {
    if (((n - 2) & i) == i)
      sum = (sum + b[i]) % 2;
  }
  cout << sum * factor;
}
### Prompt

Generate a CPP solution to 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 <cmath>
#include <iostream>
#include <vector>

using namespace std;
int main() {
  int n;
  cin >> n;
  string s;
  cin >> s;
  vector<int> b(n - 1);
  bool has_one = false;
  for (int i = 0; i < n - 1; ++i) {
    b[i] = abs(s[i + 1] - s[i]);
    if (b[i] == 1)
      has_one = true;
  }
  int factor = 1;
  if (!has_one) {
    factor = 2;
    for (int i = 0; i < n - 1; ++i)
      b[i] >>= 1;
  }
  int sum = 0;
  for (int i = 0; i < n - 1; ++i) {
    if (((n - 2) & i) == i)
      sum = (sum + b[i]) % 2;
  }
  cout << sum * factor;
}
```