#include <bits/stdc++.h>
using namespace std;
void TC() {
  long long int n;
  cin >> n;
  string s;
  cin >> s;
  long long int open = count(s.begin(), s.end(), '(');
  long long int close = n - open;
  if (abs(open - close) != 2) {
    cout << 0 << "\n";
    return;
  }
  vector<long long int> pre(n + 2, 0), suf(n + 2, 0);
  vector<bool> okp(n + 2, true), oks(n + 2, true);
  for (long long int i = 0; i < n; i++) {
    pre[i + 1] = pre[i] + (s[i] == '(' ? 1 : -1);
    okp[i + 1] = okp[i] & (pre[i + 1] >= 0);
  }
  for (long long int i = n - 1; i >= 0; i--) {
    suf[i] = suf[i + 1] + (s[i] == ')' ? 1 : -1);
    oks[i] = oks[i + 1] & (suf[i] >= 0);
  }
  long long int res = 0;
  for (long long int i = 0; i < n; i++) {
    if (!okp[i] || !oks[i + 1]) {
      continue;
    }
    if (s[i] == '(') {
      if (pre[i] - 1 >= 0 && pre[i] - 1 - suf[i + 1] == 0) {
        res++;
      }
    } else {
      if (pre[i] + 1 >= 0 && pre[i] + 1 - suf[i + 1] == 0) {
        res++;
      }
    }
  }
  cout << res << "\n";
  return;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int T = 1;
  while (T-- > 0) {
    TC();
  }
  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;
void TC() {
  long long int n;
  cin >> n;
  string s;
  cin >> s;
  long long int open = count(s.begin(), s.end(), '(');
  long long int close = n - open;
  if (abs(open - close) != 2) {
    cout << 0 << "\n";
    return;
  }
  vector<long long int> pre(n + 2, 0), suf(n + 2, 0);
  vector<bool> okp(n + 2, true), oks(n + 2, true);
  for (long long int i = 0; i < n; i++) {
    pre[i + 1] = pre[i] + (s[i] == '(' ? 1 : -1);
    okp[i + 1] = okp[i] & (pre[i + 1] >= 0);
  }
  for (long long int i = n - 1; i >= 0; i--) {
    suf[i] = suf[i + 1] + (s[i] == ')' ? 1 : -1);
    oks[i] = oks[i + 1] & (suf[i] >= 0);
  }
  long long int res = 0;
  for (long long int i = 0; i < n; i++) {
    if (!okp[i] || !oks[i + 1]) {
      continue;
    }
    if (s[i] == '(') {
      if (pre[i] - 1 >= 0 && pre[i] - 1 - suf[i + 1] == 0) {
        res++;
      }
    } else {
      if (pre[i] + 1 >= 0 && pre[i] + 1 - suf[i + 1] == 0) {
        res++;
      }
    }
  }
  cout << res << "\n";
  return;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int T = 1;
  while (T-- > 0) {
    TC();
  }
  return 0;
}

```