#include <bits/stdc++.h>
using namespace std;
const int MAX = 1e6 + 9;
int fre[MAX], mi[MAX];
int main(void) {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  string ar;
  cin >> n >> ar;
  if (n % 2 == 1)
    cout << 0 << endl;
  else {
    int now = 0;
    for (int i = 0; i < n; i++) {
      if (ar[i] == '(')
        now++;
      else
        now--;
      fre[i] = now;
    }
    mi[n - 1] = fre[n - 1];
    for (int i = n - 2; i >= 0; i--) {
      mi[i] = min(mi[i + 1], fre[i]);
    }
    int ans = 0, temp = 0;
    for (int i = 0; i < n; i++) {
      if (ar[i] == '(') {
        if (mi[i] >= 2 && now == 2) ans++;
        temp++;
      } else {
        if (mi[i] >= -2 && now == -2) ans++;
        temp--;
      }
      if (temp < 0) break;
    }
    cout << ans << endl;
  }
  return 0;
}

### 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;
const int MAX = 1e6 + 9;
int fre[MAX], mi[MAX];
int main(void) {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  string ar;
  cin >> n >> ar;
  if (n % 2 == 1)
    cout << 0 << endl;
  else {
    int now = 0;
    for (int i = 0; i < n; i++) {
      if (ar[i] == '(')
        now++;
      else
        now--;
      fre[i] = now;
    }
    mi[n - 1] = fre[n - 1];
    for (int i = n - 2; i >= 0; i--) {
      mi[i] = min(mi[i + 1], fre[i]);
    }
    int ans = 0, temp = 0;
    for (int i = 0; i < n; i++) {
      if (ar[i] == '(') {
        if (mi[i] >= 2 && now == 2) ans++;
        temp++;
      } else {
        if (mi[i] >= -2 && now == -2) ans++;
        temp--;
      }
      if (temp < 0) break;
    }
    cout << ans << endl;
  }
  return 0;
}

```