#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = pair<int, int>;
const int MOD = 1000000007;
int psum[1000006];
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = (1); i < ((n) + 1); i++) {
    char c;
    cin >> c;
    if (c == '(')
      psum[i] = 1;
    else
      psum[i] = -1;
    psum[i] += psum[i - 1];
    if (psum[i] == -3) {
      cout << 0 << endl;
      return 0;
    }
  }
  int ans = 0;
  if (psum[n] == -2)
    for (int i = (1); i < ((n) + 1); i++) {
      if (psum[i] - psum[i - 1] == -1) ans++;
      if (psum[i] == -1) break;
    }
  else if (psum[n] == 2)
    for (int i = n; i >= 1; i--) {
      if (psum[i] == 1) {
        for (int c = (1); c < ((n) + 1); c++)
          if (psum[c] == -1) ans = 0;
        break;
      }
      if (psum[i] - psum[i - 1] == 1) ans++;
    }
  cout << ans << endl;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

You are given a bracket sequence s consisting of n opening '(' and closing ')' brackets.

A regular bracket sequence is a bracket sequence that can be transformed into a correct arithmetic expression by inserting characters '1' and '+' between the original characters of the sequence. For example, bracket sequences "()()", "(())" are regular (the resulting expressions are: "(1)+(1)", "((1+1)+1)"), and ")(" and "(" are not.

You can change the type of some bracket s_i. It means that if s_i =  ')' then you can change it to '(' and vice versa.

Your task is to calculate the number of positions i such that if you change the type of the i-th bracket, then the resulting bracket sequence becomes regular.

Input

The first line of the input contains one integer n (1 ≤ n ≤ 10^6) — the length of the bracket sequence.

The second line of the input contains the string s consisting of n opening '(' and closing ')' brackets.

Output

Print one integer — the number of positions i such that if you change the type of the i-th bracket, then the resulting bracket sequence becomes regular.

Examples

Input


6
(((())


Output


3


Input


6
()()()


Output


0


Input


1
)


Output


0


Input


8
)))(((((


Output


0

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = pair<int, int>;
const int MOD = 1000000007;
int psum[1000006];
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = (1); i < ((n) + 1); i++) {
    char c;
    cin >> c;
    if (c == '(')
      psum[i] = 1;
    else
      psum[i] = -1;
    psum[i] += psum[i - 1];
    if (psum[i] == -3) {
      cout << 0 << endl;
      return 0;
    }
  }
  int ans = 0;
  if (psum[n] == -2)
    for (int i = (1); i < ((n) + 1); i++) {
      if (psum[i] - psum[i - 1] == -1) ans++;
      if (psum[i] == -1) break;
    }
  else if (psum[n] == 2)
    for (int i = n; i >= 1; i--) {
      if (psum[i] == 1) {
        for (int c = (1); c < ((n) + 1); c++)
          if (psum[c] == -1) ans = 0;
        break;
      }
      if (psum[i] - psum[i - 1] == 1) ans++;
    }
  cout << ans << endl;
}

```