#include <bits/stdc++.h>
using namespace std;
long n;
long dem = 0;
long ct[1000005], cs[1000005];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  string s;
  cin >> s;
  s = ' ' + s;
  for (long i = 1; i <= n; ++i) {
    if (ct[i - 1] < 0) {
      ct[i] = -1;
      continue;
    }
    ct[i] = ct[i - 1];
    if (s[i] == '(')
      ct[i]++;
    else
      ct[i]--;
  }
  for (long i = n; i >= 1; --i) {
    if (cs[i + 1] > 0) {
      cs[i] = 1;
      continue;
    }
    cs[i] = cs[i + 1];
    if (s[i] == '(')
      cs[i]++;
    else
      cs[i]--;
  }
  long dem = 0;
  for (long i = 1; i <= n; ++i) {
    if (ct[i - 1] >= 0 && cs[i + 1] <= 0) {
      if (s[i] == '(') {
        if (ct[i - 1] - 1 >= 0 && ct[i - 1] - 1 + cs[i + 1] == 0) {
          dem++;
        }
      } else {
        if (cs[i + 1] + 1 <= 0 && ct[i - 1] + cs[i + 1] + 1 == 0) {
          dem++;
        }
      }
    }
  }
  cout << dem;
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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 n;
long dem = 0;
long ct[1000005], cs[1000005];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  string s;
  cin >> s;
  s = ' ' + s;
  for (long i = 1; i <= n; ++i) {
    if (ct[i - 1] < 0) {
      ct[i] = -1;
      continue;
    }
    ct[i] = ct[i - 1];
    if (s[i] == '(')
      ct[i]++;
    else
      ct[i]--;
  }
  for (long i = n; i >= 1; --i) {
    if (cs[i + 1] > 0) {
      cs[i] = 1;
      continue;
    }
    cs[i] = cs[i + 1];
    if (s[i] == '(')
      cs[i]++;
    else
      cs[i]--;
  }
  long dem = 0;
  for (long i = 1; i <= n; ++i) {
    if (ct[i - 1] >= 0 && cs[i + 1] <= 0) {
      if (s[i] == '(') {
        if (ct[i - 1] - 1 >= 0 && ct[i - 1] - 1 + cs[i + 1] == 0) {
          dem++;
        }
      } else {
        if (cs[i + 1] + 1 <= 0 && ct[i - 1] + cs[i + 1] + 1 == 0) {
          dem++;
        }
      }
    }
  }
  cout << dem;
  return 0;
}

```