#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1e9 + 7;
class {
 public:
  void solve() {
    int n;
    cin >> n;
    string s;
    cin >> s;
    s = "#" + s;
    int f = 0, ans = 0;
    bool prefid = 0;
    for (int i = 1; i <= n; i++) {
      if (s[i] == '(')
        f++;
      else
        f--;
      if (f < 0 && !prefid) {
        bool ok = 0;
        for (int j = 1; j <= i; j++)
          if (s[j] == ')') {
            ans++;
            ok = 1;
          }
        if (ok)
          f += 2;
        else {
          cout << 0;
          return;
        }
        prefid = true;
      } else if (f < 0 && prefid) {
        cout << 0;
        return;
      }
    }
    if (f == 0) {
      cout << ans;
    } else if (f != 0) {
      if (prefid) {
        cout << 0;
      } else if (f == 2) {
        int nxtmin = 2;
        for (int i = n; i >= 1; i--) {
          nxtmin = min(f, nxtmin);
          if (nxtmin >= 2 && s[i] == '(') ans++;
          if (s[i] == '(')
            f--;
          else
            f++;
        }
        cout << ans;
      } else {
        cout << 0;
      }
    }
  }
} NSPACE;
int main() {
  ios_base::sync_with_stdio(false);
  cout.tie(0);
  cin.tie(0);
  NSPACE.solve();
}

### Prompt

Create a solution in Cpp for 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 = 1e9 + 7;
class {
 public:
  void solve() {
    int n;
    cin >> n;
    string s;
    cin >> s;
    s = "#" + s;
    int f = 0, ans = 0;
    bool prefid = 0;
    for (int i = 1; i <= n; i++) {
      if (s[i] == '(')
        f++;
      else
        f--;
      if (f < 0 && !prefid) {
        bool ok = 0;
        for (int j = 1; j <= i; j++)
          if (s[j] == ')') {
            ans++;
            ok = 1;
          }
        if (ok)
          f += 2;
        else {
          cout << 0;
          return;
        }
        prefid = true;
      } else if (f < 0 && prefid) {
        cout << 0;
        return;
      }
    }
    if (f == 0) {
      cout << ans;
    } else if (f != 0) {
      if (prefid) {
        cout << 0;
      } else if (f == 2) {
        int nxtmin = 2;
        for (int i = n; i >= 1; i--) {
          nxtmin = min(f, nxtmin);
          if (nxtmin >= 2 && s[i] == '(') ans++;
          if (s[i] == '(')
            f--;
          else
            f++;
        }
        cout << ans;
      } else {
        cout << 0;
      }
    }
  }
} NSPACE;
int main() {
  ios_base::sync_with_stdio(false);
  cout.tie(0);
  cin.tie(0);
  NSPACE.solve();
}

```