#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int tc = 1, cas = 1;
  while (tc--) {
    int n;
    string s;
    cin >> n >> s;
    if (n & 1)
      cout << "0" << endl;
    else {
      int len = s.length();
      int ara[len + 3];
      int last = 0;
      int minus_1 = 0, mn = 0;
      for (int i = 0; i < s.length(); i++) {
        if (s[i] == '(')
          last++;
        else
          last--;
        ara[i] = last;
        if (last == -1) minus_1++;
        mn = min(mn, last);
      }
      if (mn < -2)
        cout << 0 << endl;
      else if (last == 2) {
        if (minus_1)
          cout << 0 << endl;
        else {
          int ans = 0, ind = s.length() - 1;
          while (ind > -1) {
            if (ara[ind] == 1) break;
            if (s[ind] == '(') ans++;
            ind--;
          }
          cout << ans << endl;
        }
      } else if (last == -2) {
        int ans = 0, ind = 0;
        while (ind < s.length()) {
          if (ara[ind] == -1) {
            ans++;
            break;
          }
          if (s[ind] == ')') ans++;
          ind++;
        }
        cout << ans << endl;
      } else
        cout << 0 << endl;
    }
  }
  return 0;
}

### Prompt

Please formulate a CPP solution to the following problem:

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;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int tc = 1, cas = 1;
  while (tc--) {
    int n;
    string s;
    cin >> n >> s;
    if (n & 1)
      cout << "0" << endl;
    else {
      int len = s.length();
      int ara[len + 3];
      int last = 0;
      int minus_1 = 0, mn = 0;
      for (int i = 0; i < s.length(); i++) {
        if (s[i] == '(')
          last++;
        else
          last--;
        ara[i] = last;
        if (last == -1) minus_1++;
        mn = min(mn, last);
      }
      if (mn < -2)
        cout << 0 << endl;
      else if (last == 2) {
        if (minus_1)
          cout << 0 << endl;
        else {
          int ans = 0, ind = s.length() - 1;
          while (ind > -1) {
            if (ara[ind] == 1) break;
            if (s[ind] == '(') ans++;
            ind--;
          }
          cout << ans << endl;
        }
      } else if (last == -2) {
        int ans = 0, ind = 0;
        while (ind < s.length()) {
          if (ara[ind] == -1) {
            ans++;
            break;
          }
          if (s[ind] == ')') ans++;
          ind++;
        }
        cout << ans << endl;
      } else
        cout << 0 << endl;
    }
  }
  return 0;
}

```