#include <bits/stdc++.h>
using namespace std;
char k[1000006];
int main() {
  int n;
  cin >> n;
  cin >> k;
  int zuo = 0;
  int you = 0;
  int sum = 0;
  int ans = 0;
  for (int i = 0; i < n; i++) {
    if (k[i] == '(')
      zuo++;
    else
      you++;
  }
  if (abs(zuo - you) != 2) {
    cout << 0 << endl;
    exit(0);
  }
  int flag = 1;
  int w = 1;
  if (you > zuo) {
    int d = 0;
    for (int i = 0; i < n; i++) {
      if (k[i] == '(')
        d++;
      else {
        sum++;
        d--;
      }
      if (d == -1 && flag == 1 && w) {
        flag = 0;
        ans = sum;
      }
      if (d == -3) {
        cout << 0 << endl;
        exit(0);
      }
    }
    d = 0;
    for (int i = n - 1; i >= 0; i--) {
      if (k[i] == ')')
        d++;
      else {
        d--;
      }
      if (d == -3) {
        cout << 0 << endl;
        exit(0);
      }
    }
  } else {
    int d = 0;
    for (int i = n - 1; i >= 0; i--) {
      if (k[i] == ')')
        d++;
      else {
        sum++;
        d--;
      }
      if (d == -1 && flag == 1 && w) {
        flag = 0;
        ans = sum;
      }
      if (d == -3) {
        cout << 0 << endl;
        exit(0);
      }
    }
    d = 0;
    for (int i = 0; i < n; i++) {
      if (k[i] == '(')
        d++;
      else {
        d--;
      }
      if (d == -3) {
        cout << 0 << endl;
        exit(0);
      }
    }
  }
  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;
char k[1000006];
int main() {
  int n;
  cin >> n;
  cin >> k;
  int zuo = 0;
  int you = 0;
  int sum = 0;
  int ans = 0;
  for (int i = 0; i < n; i++) {
    if (k[i] == '(')
      zuo++;
    else
      you++;
  }
  if (abs(zuo - you) != 2) {
    cout << 0 << endl;
    exit(0);
  }
  int flag = 1;
  int w = 1;
  if (you > zuo) {
    int d = 0;
    for (int i = 0; i < n; i++) {
      if (k[i] == '(')
        d++;
      else {
        sum++;
        d--;
      }
      if (d == -1 && flag == 1 && w) {
        flag = 0;
        ans = sum;
      }
      if (d == -3) {
        cout << 0 << endl;
        exit(0);
      }
    }
    d = 0;
    for (int i = n - 1; i >= 0; i--) {
      if (k[i] == ')')
        d++;
      else {
        d--;
      }
      if (d == -3) {
        cout << 0 << endl;
        exit(0);
      }
    }
  } else {
    int d = 0;
    for (int i = n - 1; i >= 0; i--) {
      if (k[i] == ')')
        d++;
      else {
        sum++;
        d--;
      }
      if (d == -1 && flag == 1 && w) {
        flag = 0;
        ans = sum;
      }
      if (d == -3) {
        cout << 0 << endl;
        exit(0);
      }
    }
    d = 0;
    for (int i = 0; i < n; i++) {
      if (k[i] == '(')
        d++;
      else {
        d--;
      }
      if (d == -3) {
        cout << 0 << endl;
        exit(0);
      }
    }
  }
  cout << ans << endl;
  return 0;
}

```