#include <bits/stdc++.h>
using namespace std;
int n;
string s;
int prefix[1000005];
int prefix_min[1000005];
int suffix[1000005];
int suffix_min[1000005];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> s;
  s = '1' + s;
  if (n % 2) {
    cout << 0;
    return 0;
  }
  for (int i = 1; i <= n; i++) {
    if (s[i] == '(')
      prefix[i] = 1;
    else
      prefix[i] = -1;
    prefix[i] += prefix[i - 1];
    prefix_min[i] = min(prefix[i], prefix_min[i - 1]);
  }
  for (int i = n; i >= 1; i--) {
    if (s[i] == '(')
      suffix[i] = 1;
    else
      suffix[i] = -1;
    suffix[i] += suffix[i + 1];
    if (i < n) {
      if (s[i] == '(')
        suffix_min[i] = min(1, suffix_min[i + 1] + 1);
      else
        suffix_min[i] = min(-1, suffix_min[i + 1] - 1);
    } else
      suffix_min[i] = suffix[i];
  }
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    int sum = prefix[i - 1];
    if (s[i] == '(')
      sum--;
    else
      sum++;
    int mn = min(prefix_min[i - 1], sum);
    if (mn < 0) continue;
    if (sum + suffix[i + 1] == 0 && sum + suffix_min[i + 1] >= 0) {
      ans++;
    }
  }
  cout << ans;
}

### 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;
int n;
string s;
int prefix[1000005];
int prefix_min[1000005];
int suffix[1000005];
int suffix_min[1000005];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> s;
  s = '1' + s;
  if (n % 2) {
    cout << 0;
    return 0;
  }
  for (int i = 1; i <= n; i++) {
    if (s[i] == '(')
      prefix[i] = 1;
    else
      prefix[i] = -1;
    prefix[i] += prefix[i - 1];
    prefix_min[i] = min(prefix[i], prefix_min[i - 1]);
  }
  for (int i = n; i >= 1; i--) {
    if (s[i] == '(')
      suffix[i] = 1;
    else
      suffix[i] = -1;
    suffix[i] += suffix[i + 1];
    if (i < n) {
      if (s[i] == '(')
        suffix_min[i] = min(1, suffix_min[i + 1] + 1);
      else
        suffix_min[i] = min(-1, suffix_min[i + 1] - 1);
    } else
      suffix_min[i] = suffix[i];
  }
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    int sum = prefix[i - 1];
    if (s[i] == '(')
      sum--;
    else
      sum++;
    int mn = min(prefix_min[i - 1], sum);
    if (mn < 0) continue;
    if (sum + suffix[i + 1] == 0 && sum + suffix_min[i + 1] >= 0) {
      ans++;
    }
  }
  cout << ans;
}

```