#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e6;
const long long MAX = 1e6 + 10;
const long long inf = 1e18 + 7;
long long a[MAX];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n, ans = 0;
  cin >> n;
  if (n & 1) {
    cout << "0\n";
    return 0;
  }
  string s;
  cin >> s;
  for (long long i = 0; i < n; i++) {
    if (i) a[i] = a[i - 1];
    if (s[i] == '(')
      a[i]++;
    else
      a[i]--;
    if (a[i] < -2) {
      cout << "0\n";
      return 0;
    }
  }
  if (a[n - 1] != 2 && a[n - 1] != -2) {
    cout << "0\n";
    return 0;
  }
  if (a[n - 1] == 2) {
    for (long long i = 0; i < n; i++) {
      if (a[i] < 0) {
        cout << "0\n";
        return 0;
      }
    }
    for (long long i = n - 1; i >= 0; i--) {
      if (a[i] < 2) break;
      if (s[i] == '(') ans++;
    }
  }
  if (a[n - 1] == -2) {
    for (long long i = 0; i < n; i++) {
      if (i) {
        if (a[i - 1] < 0) break;
      }
      if (s[i] == ')') ans++;
    }
  }
  cout << ans << "\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;
const long long mod = 1e6;
const long long MAX = 1e6 + 10;
const long long inf = 1e18 + 7;
long long a[MAX];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n, ans = 0;
  cin >> n;
  if (n & 1) {
    cout << "0\n";
    return 0;
  }
  string s;
  cin >> s;
  for (long long i = 0; i < n; i++) {
    if (i) a[i] = a[i - 1];
    if (s[i] == '(')
      a[i]++;
    else
      a[i]--;
    if (a[i] < -2) {
      cout << "0\n";
      return 0;
    }
  }
  if (a[n - 1] != 2 && a[n - 1] != -2) {
    cout << "0\n";
    return 0;
  }
  if (a[n - 1] == 2) {
    for (long long i = 0; i < n; i++) {
      if (a[i] < 0) {
        cout << "0\n";
        return 0;
      }
    }
    for (long long i = n - 1; i >= 0; i--) {
      if (a[i] < 2) break;
      if (s[i] == '(') ans++;
    }
  }
  if (a[n - 1] == -2) {
    for (long long i = 0; i < n; i++) {
      if (i) {
        if (a[i - 1] < 0) break;
      }
      if (s[i] == ')') ans++;
    }
  }
  cout << ans << "\n";
  return 0;
}

```