#include <bits/stdc++.h>
using namespace std;
const int maxM = 1000000;
string cad;
vector<int> izq(maxM), der(maxM);
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  int n, ans, i;
  cin >> n;
  cin >> cad;
  izq[0] = 0;
  if (cad[0] == '(')
    izq[0]++;
  else
    izq[0]--;
  for (i = 1; i < n; i++) {
    izq[i] = izq[i - 1];
    if (cad[i] == '(')
      izq[i]++;
    else
      izq[i]--;
  }
  der[n - 1] = 0;
  if (cad[n - 1] == '(')
    der[n - 1]++;
  else
    der[n - 1]--;
  for (i = n - 2; i >= 0; i--) {
    der[i] = der[i + 1];
    if (cad[i] == '(')
      der[i]++;
    else
      der[i]--;
  }
  ans = 0;
  if (izq[n - 1] == 2) {
    for (i = n - 1; i >= 0; i--) {
      if (der[i] >= 3) break;
    }
    if (i == -1) {
      for (i = n - 1; i >= 0; i--) {
        if (der[i] <= 1) {
          if (cad[i] == '(') ans++;
        }
        if (der[i] > 0) break;
      }
    }
  } else if (izq[n - 1] == -2) {
    for (i = 0; i < n; i++) {
      if (izq[i] <= -3) break;
    }
    if (i == n) {
      for (i = 0; i < n; i++) {
        if (izq[i] >= -1) {
          if (cad[i] == ')') ans++;
        }
        if (izq[i] < 0) break;
      }
    }
  }
  if (ans == -1) ans = 0;
  cout << ans << '\n';
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 maxM = 1000000;
string cad;
vector<int> izq(maxM), der(maxM);
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  int n, ans, i;
  cin >> n;
  cin >> cad;
  izq[0] = 0;
  if (cad[0] == '(')
    izq[0]++;
  else
    izq[0]--;
  for (i = 1; i < n; i++) {
    izq[i] = izq[i - 1];
    if (cad[i] == '(')
      izq[i]++;
    else
      izq[i]--;
  }
  der[n - 1] = 0;
  if (cad[n - 1] == '(')
    der[n - 1]++;
  else
    der[n - 1]--;
  for (i = n - 2; i >= 0; i--) {
    der[i] = der[i + 1];
    if (cad[i] == '(')
      der[i]++;
    else
      der[i]--;
  }
  ans = 0;
  if (izq[n - 1] == 2) {
    for (i = n - 1; i >= 0; i--) {
      if (der[i] >= 3) break;
    }
    if (i == -1) {
      for (i = n - 1; i >= 0; i--) {
        if (der[i] <= 1) {
          if (cad[i] == '(') ans++;
        }
        if (der[i] > 0) break;
      }
    }
  } else if (izq[n - 1] == -2) {
    for (i = 0; i < n; i++) {
      if (izq[i] <= -3) break;
    }
    if (i == n) {
      for (i = 0; i < n; i++) {
        if (izq[i] >= -1) {
          if (cad[i] == ')') ans++;
        }
        if (izq[i] < 0) break;
      }
    }
  }
  if (ans == -1) ans = 0;
  cout << ans << '\n';
}

```