#include <bits/stdc++.h>
using namespace std;
int diff[1000005], l, r, n;
void does(char str[]) {
  for (int i = 0; i < n; i++) {
    if (diff[i] < 0) {
      printf("0\n");
      return;
    }
  }
  if (str[n - 1] == '(') {
    printf("1\n");
    return;
  }
  int now;
  for (int i = n - 1; i >= 0; i--) {
    if (diff[i] < 2) {
      now = i + 1;
      break;
    }
  }
  int number = 0;
  for (int i = now; i < n; i++) {
    if (str[i] == '(') {
      number++;
    }
  }
  printf("%d\n", number);
}
int main() {
  int now;
  l = r = now = 0;
  char bracket[1000005], bracket2[1000005];
  scanf("%d%s", &n, &bracket);
  bracket2[n] = '\0';
  for (int i = 0; i < n; i++) {
    bracket2[i] = (bracket[n - i - 1] == '(') ? ')' : '(';
    if (bracket[i] == '(') {
      l++;
      now++;
      diff[i] = now;
    } else {
      r++;
      now--;
      diff[i] = now;
    }
  }
  if (l - r == 2) {
    does(bracket);
  } else if (r - l == 2) {
    l = r = now = 0;
    for (int i = 0; i < n; i++) {
      if (bracket2[i] == '(') {
        l++;
        now++;
        diff[i] = now;
      } else {
        r++;
        now--;
        diff[i] = now;
      }
    }
    does(bracket2);
  } else {
    printf("0\n");
  }
  return 0;
}

### Prompt

Your task is to create a CPP solution to 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 diff[1000005], l, r, n;
void does(char str[]) {
  for (int i = 0; i < n; i++) {
    if (diff[i] < 0) {
      printf("0\n");
      return;
    }
  }
  if (str[n - 1] == '(') {
    printf("1\n");
    return;
  }
  int now;
  for (int i = n - 1; i >= 0; i--) {
    if (diff[i] < 2) {
      now = i + 1;
      break;
    }
  }
  int number = 0;
  for (int i = now; i < n; i++) {
    if (str[i] == '(') {
      number++;
    }
  }
  printf("%d\n", number);
}
int main() {
  int now;
  l = r = now = 0;
  char bracket[1000005], bracket2[1000005];
  scanf("%d%s", &n, &bracket);
  bracket2[n] = '\0';
  for (int i = 0; i < n; i++) {
    bracket2[i] = (bracket[n - i - 1] == '(') ? ')' : '(';
    if (bracket[i] == '(') {
      l++;
      now++;
      diff[i] = now;
    } else {
      r++;
      now--;
      diff[i] = now;
    }
  }
  if (l - r == 2) {
    does(bracket);
  } else if (r - l == 2) {
    l = r = now = 0;
    for (int i = 0; i < n; i++) {
      if (bracket2[i] == '(') {
        l++;
        now++;
        diff[i] = now;
      } else {
        r++;
        now--;
        diff[i] = now;
      }
    }
    does(bracket2);
  } else {
    printf("0\n");
  }
  return 0;
}

```