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

int main() {
  cin.tie(nullptr);
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  vector<int> a(n);
  for (auto&& e : a) {
    char c;
    cin >> c;
    e = c - '1';
  }
  bool f = false;
  if (count(begin(a), end(a), 1) == 0) {
    f = true;
    for (auto&& e : a) {
      e /= 2;
    }
  }
  int res = 0;
  for (int i = 0; i < n; ++i) {
    int j = n - i - 1;
    if ((i & j) == 0) {
      res += a[i];
    }
    res &= 1;
  }
  if (f) {
    res *= 2;
  }
  cout << res << '\n';
}

### 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;

int main() {
  cin.tie(nullptr);
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  vector<int> a(n);
  for (auto&& e : a) {
    char c;
    cin >> c;
    e = c - '1';
  }
  bool f = false;
  if (count(begin(a), end(a), 1) == 0) {
    f = true;
    for (auto&& e : a) {
      e /= 2;
    }
  }
  int res = 0;
  for (int i = 0; i < n; ++i) {
    int j = n - i - 1;
    if ((i & j) == 0) {
      res += a[i];
    }
    res &= 1;
  }
  if (f) {
    res *= 2;
  }
  cout << res << '\n';
}

```