#include <bits/stdc++.h>
using namespace std;
int a[int(2e6)], b[int(2e6)];
int main() {
  int n;
  cin >> n;
  if (n % 2 == 1) cout << 0, exit(0);
  string s;
  cin >> s;
  int l = n + 1, r = 0;
  for (int i = 1; i <= n; ++i) {
    if (s[i - 1] == '(')
      a[i] = a[i - 1] + 1;
    else
      a[i] = a[i - 1] - 1;
    if (a[i] < -2) cout << "0", exit(0);
  }
  int ans = 0;
  for (int i = n; i > 0; --i) {
    if (s[i - 1] == ')')
      b[i] = b[i + 1] + 1;
    else
      b[i] = b[i + 1] - 1;
    if (b[i] < -2) cout << "0", exit(0);
  }
  for (int i = n; i > 0; --i) {
    if (b[i + 1] >= 0) {
      if (s[i - 1] == '(' && a[i] - 2 >= 0 && a[i] - 2 == b[i] + 1) ans++;
      if (s[i - 1] == ')' && b[i] - 2 >= 0 && b[i] - 2 == a[i] + 1) ans++;
    } else
      break;
  }
  int bns = 0;
  for (int i = 1; i <= n; ++i) {
    if (a[i - 1] >= 0) {
      if (s[i - 1] == '(' && a[i] - 2 >= 0 && a[i] - 2 == b[i] + 1) bns++;
      if (s[i - 1] == ')' && b[i] - 2 >= 0 && b[i] - 2 == a[i] + 1) bns++;
    } else
      break;
  }
  cout << min(ans, bns);
  return 0;
}

### Prompt

In Cpp, your task is to solve 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 a[int(2e6)], b[int(2e6)];
int main() {
  int n;
  cin >> n;
  if (n % 2 == 1) cout << 0, exit(0);
  string s;
  cin >> s;
  int l = n + 1, r = 0;
  for (int i = 1; i <= n; ++i) {
    if (s[i - 1] == '(')
      a[i] = a[i - 1] + 1;
    else
      a[i] = a[i - 1] - 1;
    if (a[i] < -2) cout << "0", exit(0);
  }
  int ans = 0;
  for (int i = n; i > 0; --i) {
    if (s[i - 1] == ')')
      b[i] = b[i + 1] + 1;
    else
      b[i] = b[i + 1] - 1;
    if (b[i] < -2) cout << "0", exit(0);
  }
  for (int i = n; i > 0; --i) {
    if (b[i + 1] >= 0) {
      if (s[i - 1] == '(' && a[i] - 2 >= 0 && a[i] - 2 == b[i] + 1) ans++;
      if (s[i - 1] == ')' && b[i] - 2 >= 0 && b[i] - 2 == a[i] + 1) ans++;
    } else
      break;
  }
  int bns = 0;
  for (int i = 1; i <= n; ++i) {
    if (a[i - 1] >= 0) {
      if (s[i - 1] == '(' && a[i] - 2 >= 0 && a[i] - 2 == b[i] + 1) bns++;
      if (s[i - 1] == ')' && b[i] - 2 >= 0 && b[i] - 2 == a[i] + 1) bns++;
    } else
      break;
  }
  cout << min(ans, bns);
  return 0;
}

```