#include <bits/stdc++.h>
using namespace std;
const int N = (int)1e5;
const int INF = (int)1e9 + 7;
int main(int argc, char *argv[]) {
  int n;
  scanf("%d", &n);
  char s[n + 7];
  scanf("%s", s);
  int x = 0, y = 0;
  int a[n + 7];
  int b[n + 7];
  a[0] = b[n + 1] = 0;
  for (int i = 0; i < n; ++i) {
    if (s[i] == '(') {
      ++x;
    } else {
      if (x < 1)
        ++y;
      else
        --x;
    }
    a[i + 1] = x;
  }
  x = y = 0;
  for (int i = n - 1; i >= 0; --i) {
    if (s[i] == ')') {
      ++x;
    } else {
      if (x < 1)
        ++y;
      else
        --x;
    }
    b[i + 1] = x;
  }
  if (n < 2 || (x > 0 && y > 0) || abs(x - y) != 2) {
    puts("0");
  } else {
    int ans = 0;
    if (x > y) {
      for (int i = n; i >= 0; --i) {
        if (s[i - 1] == ')') {
          if (b[i] - 1 - (a[i - 1] + 1) == 0) ++ans;
        }
      }
    } else {
      for (int i = 1; i <= n; ++i) {
        if (s[i - 1] == '(') {
          if (a[i] - 1 - (b[i + 1] + 1) == 0) ++ans;
        }
      }
    }
    printf("%d\n", ans);
  }
  return 0;
}

### Prompt

Please formulate 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;
const int N = (int)1e5;
const int INF = (int)1e9 + 7;
int main(int argc, char *argv[]) {
  int n;
  scanf("%d", &n);
  char s[n + 7];
  scanf("%s", s);
  int x = 0, y = 0;
  int a[n + 7];
  int b[n + 7];
  a[0] = b[n + 1] = 0;
  for (int i = 0; i < n; ++i) {
    if (s[i] == '(') {
      ++x;
    } else {
      if (x < 1)
        ++y;
      else
        --x;
    }
    a[i + 1] = x;
  }
  x = y = 0;
  for (int i = n - 1; i >= 0; --i) {
    if (s[i] == ')') {
      ++x;
    } else {
      if (x < 1)
        ++y;
      else
        --x;
    }
    b[i + 1] = x;
  }
  if (n < 2 || (x > 0 && y > 0) || abs(x - y) != 2) {
    puts("0");
  } else {
    int ans = 0;
    if (x > y) {
      for (int i = n; i >= 0; --i) {
        if (s[i - 1] == ')') {
          if (b[i] - 1 - (a[i - 1] + 1) == 0) ++ans;
        }
      }
    } else {
      for (int i = 1; i <= n; ++i) {
        if (s[i - 1] == '(') {
          if (a[i] - 1 - (b[i + 1] + 1) == 0) ++ans;
        }
      }
    }
    printf("%d\n", ans);
  }
  return 0;
}

```