#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cin.tie(0), cout.tie(0),
      cout.precision(15);
  int n;
  string s;
  cin >> n >> s;
  vector<int> sum(n + 1), mn(n + 1);
  for (int i = 1; i <= n; i++)
    sum[i] = sum[i - 1] + ((s[i - 1] == '(') ? 1 : -1);
  mn[n] = sum[n];
  for (int i = n - 1; i; --i) mn[i] = min(mn[i + 1], sum[i]);
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    int t = ((s[i - 1] == '(') ? -1 : 1);
    t *= 2;
    if (mn[i] + t >= 0 && sum[n] + t == 0) ++ans;
    if (sum[i] < 0) break;
  }
  cout << ans;
  return 0;
}

### Prompt

Please create a solution in CPP 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;
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cin.tie(0), cout.tie(0),
      cout.precision(15);
  int n;
  string s;
  cin >> n >> s;
  vector<int> sum(n + 1), mn(n + 1);
  for (int i = 1; i <= n; i++)
    sum[i] = sum[i - 1] + ((s[i - 1] == '(') ? 1 : -1);
  mn[n] = sum[n];
  for (int i = n - 1; i; --i) mn[i] = min(mn[i + 1], sum[i]);
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    int t = ((s[i - 1] == '(') ? -1 : 1);
    t *= 2;
    if (mn[i] + t >= 0 && sum[n] + t == 0) ++ans;
    if (sum[i] < 0) break;
  }
  cout << ans;
  return 0;
}

```