#include <bits/stdc++.h>
using namespace std;
const int mx = 1e6 + 10;
char s[mx];
int sum[mx];
int main() {
  int n;
  scanf("%d%s", &n, s + 1);
  for (int i = 1; i <= n; i++) {
    if (s[i] == '(')
      sum[i] = sum[i - 1] + 1;
    else
      sum[i] = sum[i - 1] - 1;
  }
  if (sum[n] != 2 && sum[n] != -2) {
    printf("0\n");
    return 0;
  }
  int ans = 0;
  if (sum[n] == 2) {
    int k = 0;
    for (int i = 1; i <= n; i++) {
      if (sum[i] < 0) {
        k = i;
        break;
      }
    }
    for (int i = n; i >= 1; i--) {
      if (sum[i] < 2) break;
      if (s[i] == '(') ans++;
    }
    if (k != 0)
      printf("0\n");
    else
      printf("%d\n", ans);
  } else {
    bool ok = 1;
    for (int i = 1; i <= n; i++) {
      if (sum[i] < -2) {
        ok = 0;
        break;
      }
    }
    int k = 0;
    for (int i = 1; i <= n; i++) {
      if (sum[i] < 0) {
        k = i;
        break;
      }
    }
    for (int i = 1; i <= k; i++) {
      if (s[i] == ')') ans++;
    }
    if (!ok)
      printf("0\n");
    else
      printf("%d\n", ans);
  }
  return 0;
}

### Prompt

Please provide a CPP coded solution 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>
using namespace std;
const int mx = 1e6 + 10;
char s[mx];
int sum[mx];
int main() {
  int n;
  scanf("%d%s", &n, s + 1);
  for (int i = 1; i <= n; i++) {
    if (s[i] == '(')
      sum[i] = sum[i - 1] + 1;
    else
      sum[i] = sum[i - 1] - 1;
  }
  if (sum[n] != 2 && sum[n] != -2) {
    printf("0\n");
    return 0;
  }
  int ans = 0;
  if (sum[n] == 2) {
    int k = 0;
    for (int i = 1; i <= n; i++) {
      if (sum[i] < 0) {
        k = i;
        break;
      }
    }
    for (int i = n; i >= 1; i--) {
      if (sum[i] < 2) break;
      if (s[i] == '(') ans++;
    }
    if (k != 0)
      printf("0\n");
    else
      printf("%d\n", ans);
  } else {
    bool ok = 1;
    for (int i = 1; i <= n; i++) {
      if (sum[i] < -2) {
        ok = 0;
        break;
      }
    }
    int k = 0;
    for (int i = 1; i <= n; i++) {
      if (sum[i] < 0) {
        k = i;
        break;
      }
    }
    for (int i = 1; i <= k; i++) {
      if (s[i] == ')') ans++;
    }
    if (!ok)
      printf("0\n");
    else
      printf("%d\n", ans);
  }
  return 0;
}

```