#include <bits/stdc++.h>
int n;
bool s[1000005];
int a[1000005];
int main() {
  scanf("%d", &n);
  int i;
  for (i = 1; i <= n; i++) {
    char data;
    scanf(" %c", &data);
    if (data == '(') s[i] = 1;
  }
  int type = 0, err;
  for (i = 1; i <= n; i++) {
    if (s[i])
      a[i] = a[i - 1] + 1;
    else
      a[i] = a[i - 1] - 1;
    if (a[i] < 0) {
      type = 1;
      err = i;
      break;
    }
  }
  int ans = 0;
  if (type == 1) {
    a[err] += 2;
    for (i = err + 1; i <= n; i++) {
      if (s[i])
        a[i] = a[i - 1] + 1;
      else
        a[i] = a[i - 1] - 1;
      if (a[i] < 0) {
        printf("0");
        return 0;
      }
    }
    if (a[n] > 0) {
      printf("0");
      return 0;
    }
    for (i = err; i >= 1; i--) {
      if (!s[i]) ans++;
    }
    printf("%d", ans);
    return 0;
  }
  if (a[n] != 2) {
    printf("0");
    return 0;
  }
  err = 0;
  for (i = n; i >= 1; i--) {
    a[i] -= 2;
    if (a[i] < 0) break;
    if (s[i]) ans++;
  }
  printf("%d", ans);
  return 0;
}

### Prompt

Develop a solution in CPP 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>
int n;
bool s[1000005];
int a[1000005];
int main() {
  scanf("%d", &n);
  int i;
  for (i = 1; i <= n; i++) {
    char data;
    scanf(" %c", &data);
    if (data == '(') s[i] = 1;
  }
  int type = 0, err;
  for (i = 1; i <= n; i++) {
    if (s[i])
      a[i] = a[i - 1] + 1;
    else
      a[i] = a[i - 1] - 1;
    if (a[i] < 0) {
      type = 1;
      err = i;
      break;
    }
  }
  int ans = 0;
  if (type == 1) {
    a[err] += 2;
    for (i = err + 1; i <= n; i++) {
      if (s[i])
        a[i] = a[i - 1] + 1;
      else
        a[i] = a[i - 1] - 1;
      if (a[i] < 0) {
        printf("0");
        return 0;
      }
    }
    if (a[n] > 0) {
      printf("0");
      return 0;
    }
    for (i = err; i >= 1; i--) {
      if (!s[i]) ans++;
    }
    printf("%d", ans);
    return 0;
  }
  if (a[n] != 2) {
    printf("0");
    return 0;
  }
  err = 0;
  for (i = n; i >= 1; i--) {
    a[i] -= 2;
    if (a[i] < 0) break;
    if (s[i]) ans++;
  }
  printf("%d", ans);
  return 0;
}

```