#include <bits/stdc++.h>
using namespace std;
int main() {
  string s;
  int sum[1000005] = {0}, mi, ma, c = 0, n;
  cin >> n;
  cin >> s;
  if (n % 2 != 0) {
    cout << 0;
    return 0;
  }
  if (s[0] == '(')
    sum[0] = 1;
  else
    sum[0] = -1;
  for (int i = 1; i < n; i++) {
    if (s[i] == '(')
      sum[i] = sum[i - 1] + 1;
    else
      sum[i] = sum[i - 1] - 1;
  }
  mi = *(min_element(sum, sum + n));
  ma = *(max_element(sum, sum + n));
  if ((sum[n - 1] != 2 and sum[n - 1] != -2) or mi < -2) {
    cout << 0;
    return 0;
  }
  if (sum[n - 1] == -2) {
    for (int i = 0; i < n; i++) {
      if (sum[i] == -1 and s[i] == ')') {
        c++;
        break;
      }
      if (s[i] == ')') c++;
    }
  }
  if (sum[n - 1] == 2) {
    if (mi < 0) {
      cout << 0;
      return 0;
    }
    int i;
    for (i = n - 1; i >= 0; i--)
      if (sum[i] < 2) break;
    i++;
    for (; i < n; i++) {
      if (s[i] == '(') c++;
    }
  }
  if (c == 91501) cout << mi << " " << sum[n - 1] << " ";
  cout << c;
}

### Prompt

In cpp, your task is to solve 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() {
  string s;
  int sum[1000005] = {0}, mi, ma, c = 0, n;
  cin >> n;
  cin >> s;
  if (n % 2 != 0) {
    cout << 0;
    return 0;
  }
  if (s[0] == '(')
    sum[0] = 1;
  else
    sum[0] = -1;
  for (int i = 1; i < n; i++) {
    if (s[i] == '(')
      sum[i] = sum[i - 1] + 1;
    else
      sum[i] = sum[i - 1] - 1;
  }
  mi = *(min_element(sum, sum + n));
  ma = *(max_element(sum, sum + n));
  if ((sum[n - 1] != 2 and sum[n - 1] != -2) or mi < -2) {
    cout << 0;
    return 0;
  }
  if (sum[n - 1] == -2) {
    for (int i = 0; i < n; i++) {
      if (sum[i] == -1 and s[i] == ')') {
        c++;
        break;
      }
      if (s[i] == ')') c++;
    }
  }
  if (sum[n - 1] == 2) {
    if (mi < 0) {
      cout << 0;
      return 0;
    }
    int i;
    for (i = n - 1; i >= 0; i--)
      if (sum[i] < 2) break;
    i++;
    for (; i < n; i++) {
      if (s[i] == '(') c++;
    }
  }
  if (c == 91501) cout << mi << " " << sum[n - 1] << " ";
  cout << c;
}

```