#include <bits/stdc++.h>
using namespace std;
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long poww(long long a, int b) {
  long long ans = 1;
  for (int i = 1; i <= b; i++) ans *= a;
  return ans;
}
int x[1000005];
int last[1000005];
int first[1000005];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int ans = 0, n;
  string s;
  cin >> n;
  cin >> s;
  for (int i = 0; i < n; i++) {
    if (i != 0) x[i] = x[i - 1];
    x[i] += (s[i] == '(') - (s[i] == ')');
  }
  last[n - 1] = x[n - 1];
  for (int i = n - 2; i >= 0; i--) last[i] = min(last[i + 1], x[i]);
  first[0] = x[0];
  for (int i = 1; i < n; i++) first[i] = min(first[i - 1], x[i]);
  if (x[n - 1] == 2) {
    for (int i = 1; i < n; i++) {
      if (s[i] == '(' && last[i] == 2 && first[i - 1] >= 0) {
        ans++;
      }
    }
    cout << ans;
  } else if (x[n - 1] == -2) {
    for (int i = 0; i < n - 1; i++) {
      if (s[i] == ')' && last[i] == -2) {
        if (i == 0)
          ans++;
        else if ((i > 0 && first[i - 1] >= 0))
          ans++;
      }
    }
    cout << ans;
  } else
    cout << "0";
}

### Prompt

Please formulate 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;
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long poww(long long a, int b) {
  long long ans = 1;
  for (int i = 1; i <= b; i++) ans *= a;
  return ans;
}
int x[1000005];
int last[1000005];
int first[1000005];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int ans = 0, n;
  string s;
  cin >> n;
  cin >> s;
  for (int i = 0; i < n; i++) {
    if (i != 0) x[i] = x[i - 1];
    x[i] += (s[i] == '(') - (s[i] == ')');
  }
  last[n - 1] = x[n - 1];
  for (int i = n - 2; i >= 0; i--) last[i] = min(last[i + 1], x[i]);
  first[0] = x[0];
  for (int i = 1; i < n; i++) first[i] = min(first[i - 1], x[i]);
  if (x[n - 1] == 2) {
    for (int i = 1; i < n; i++) {
      if (s[i] == '(' && last[i] == 2 && first[i - 1] >= 0) {
        ans++;
      }
    }
    cout << ans;
  } else if (x[n - 1] == -2) {
    for (int i = 0; i < n - 1; i++) {
      if (s[i] == ')' && last[i] == -2) {
        if (i == 0)
          ans++;
        else if ((i > 0 && first[i - 1] >= 0))
          ans++;
      }
    }
    cout << ans;
  } else
    cout << "0";
}

```