#include <bits/stdc++.h>
using namespace std;
int p1[1000002];
int s1[1000002];
bool p[1000002];
bool s[1000002];
int main() {
  int n;
  string q;
  cin >> n;
  cin >> q;
  int k1 = 0;
  int k2 = 0;
  if (q[0] == '(') {
    p1[0] = 1;
    p[0] = 1;
  } else {
    p1[0] = -1;
    p[0] = 0;
  }
  for (int i = 1; i < q.length(); i++) {
    if (q[i] == '(') {
      p1[i] = p1[i - 1] + 1;
    } else {
      p1[i] = p1[i - 1] - 1;
    }
    if (p[i - 1] && p1[i] >= 0) {
      p[i] = 1;
    }
  }
  if (q[q.length() - 1] == ')') {
    s1[q.length() - 1] = -1;
    s[q.length() - 1] = 1;
  } else {
    s1[q.length() - 1] = 1;
    s[q.length() - 1] = 0;
  }
  for (int i = q.length() - 2; i >= 0; i--) {
    if (q[i] == '(') {
      s1[i] = s1[i + 1] + 1;
    } else {
      s1[i] = s1[i + 1] - 1;
    }
    if (s[i + 1] && s1[i] <= 0) {
      s[i] = 1;
    }
  }
  int res = 0;
  for (int i = 0; i < q.length(); i++) {
  }
  for (int i = 0; i < q.length(); i++) {
  }
  for (int i = 0; i < q.length(); i++) {
  }
  for (int i = 0; i < q.length(); i++) {
  }
  if (q[0] == ')') {
    if (n > 2) {
      if (1 + s1[1] == 0 && s[2]) {
        res++;
      }
    } else {
      if (1 + s1[1] == 0) {
        res++;
      }
    }
  }
  if (q[q.length() - 1] == '(' && p[q.length() - 2]) {
    if (p1[q.length() - 2] - 1 == 0) {
      res++;
    }
  }
  for (int i = 1; i < q.length() - 1; i++) {
    if (p[i - 1] && s[i + 1]) {
      if (q[i] == '(') {
        if (p1[i - 1] > 0 && p1[i - 1] - 1 + s1[i + 1] == 0) {
          res++;
        }
      }
      if (q[i] == ')') {
        if (p1[i - 1] + 1 + s1[i + 1] == 0) {
          res++;
        }
      }
    }
  }
  cout << res << endl;
  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 p1[1000002];
int s1[1000002];
bool p[1000002];
bool s[1000002];
int main() {
  int n;
  string q;
  cin >> n;
  cin >> q;
  int k1 = 0;
  int k2 = 0;
  if (q[0] == '(') {
    p1[0] = 1;
    p[0] = 1;
  } else {
    p1[0] = -1;
    p[0] = 0;
  }
  for (int i = 1; i < q.length(); i++) {
    if (q[i] == '(') {
      p1[i] = p1[i - 1] + 1;
    } else {
      p1[i] = p1[i - 1] - 1;
    }
    if (p[i - 1] && p1[i] >= 0) {
      p[i] = 1;
    }
  }
  if (q[q.length() - 1] == ')') {
    s1[q.length() - 1] = -1;
    s[q.length() - 1] = 1;
  } else {
    s1[q.length() - 1] = 1;
    s[q.length() - 1] = 0;
  }
  for (int i = q.length() - 2; i >= 0; i--) {
    if (q[i] == '(') {
      s1[i] = s1[i + 1] + 1;
    } else {
      s1[i] = s1[i + 1] - 1;
    }
    if (s[i + 1] && s1[i] <= 0) {
      s[i] = 1;
    }
  }
  int res = 0;
  for (int i = 0; i < q.length(); i++) {
  }
  for (int i = 0; i < q.length(); i++) {
  }
  for (int i = 0; i < q.length(); i++) {
  }
  for (int i = 0; i < q.length(); i++) {
  }
  if (q[0] == ')') {
    if (n > 2) {
      if (1 + s1[1] == 0 && s[2]) {
        res++;
      }
    } else {
      if (1 + s1[1] == 0) {
        res++;
      }
    }
  }
  if (q[q.length() - 1] == '(' && p[q.length() - 2]) {
    if (p1[q.length() - 2] - 1 == 0) {
      res++;
    }
  }
  for (int i = 1; i < q.length() - 1; i++) {
    if (p[i - 1] && s[i + 1]) {
      if (q[i] == '(') {
        if (p1[i - 1] > 0 && p1[i - 1] - 1 + s1[i + 1] == 0) {
          res++;
        }
      }
      if (q[i] == ')') {
        if (p1[i - 1] + 1 + s1[i + 1] == 0) {
          res++;
        }
      }
    }
  }
  cout << res << endl;
  return 0;
}

```