#include <bits/stdc++.h>
using namespace std;
int main() {
  long int n, i, o = 0, c = 0, count = 0, sum = 0;
  cin >> n;
  char s[n];
  cin >> s;
  long int arr[n];
  for (i = 0; i < n; i++) {
    if (s[i] == ')')
      c++;
    else
      o++;
  }
  if (o - c == -2) {
    reverse(s, s + n);
    for (i = 0; i < n; i++) {
      if (s[i] == ')')
        s[i] = '(';
      else
        s[i] = ')';
    }
  }
  for (i = 0; i < n; i++) {
    if (s[i] == '(') {
      sum++;
      arr[i] = sum;
      if (arr[i] < 0) {
        cout << 0;
        return 0;
      }
    }
    if (s[i] == ')') {
      sum--;
      arr[i] = sum;
      if (arr[i] < 0) {
        cout << 0;
        return 0;
      }
    }
  }
  if (sum == 2 && s[n - 1] == ')') {
    for (i = 0; i < n; i++) {
      if (arr[i] > 1 && s[i] == '(') count++;
      if (arr[i] <= 1 && s[i] == ')') count = 0;
    }
  }
  if (sum == 2 && s[n - 1] == '(') count = 1;
  cout << count;
  return 0;
}

### Prompt

Create a solution in Cpp for 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() {
  long int n, i, o = 0, c = 0, count = 0, sum = 0;
  cin >> n;
  char s[n];
  cin >> s;
  long int arr[n];
  for (i = 0; i < n; i++) {
    if (s[i] == ')')
      c++;
    else
      o++;
  }
  if (o - c == -2) {
    reverse(s, s + n);
    for (i = 0; i < n; i++) {
      if (s[i] == ')')
        s[i] = '(';
      else
        s[i] = ')';
    }
  }
  for (i = 0; i < n; i++) {
    if (s[i] == '(') {
      sum++;
      arr[i] = sum;
      if (arr[i] < 0) {
        cout << 0;
        return 0;
      }
    }
    if (s[i] == ')') {
      sum--;
      arr[i] = sum;
      if (arr[i] < 0) {
        cout << 0;
        return 0;
      }
    }
  }
  if (sum == 2 && s[n - 1] == ')') {
    for (i = 0; i < n; i++) {
      if (arr[i] > 1 && s[i] == '(') count++;
      if (arr[i] <= 1 && s[i] == ')') count = 0;
    }
  }
  if (sum == 2 && s[n - 1] == '(') count = 1;
  cout << count;
  return 0;
}

```