#include <bits/stdc++.h>
using namespace std;
long long a[1000100][2], b[1000100][2];
int32_t main() {
  long long n;
  cin >> n;
  string s1;
  cin >> s1;
  stack<char> s;
  long long j = 1;
  for (char i : s1) {
    if (i == '(')
      a[j][1] = a[j - 1][1] + 1, a[j][0] = a[j - 1][0];
    else {
      if (a[j - 1][1] > 0)
        a[j][1] = a[j - 1][1] - 1, a[j][0] = a[j - 1][0];
      else
        a[j][0] = a[j - 1][0] + 1, a[j][1] = a[j - 1][1];
    }
    j++;
  }
  j = n;
  string s2 = s1;
  reverse(s2.begin(), s2.end());
  for (char i : s2) {
    if (i == ')')
      b[j][0] = b[j + 1][0] + 1, b[j][1] = b[j + 1][1];
    else {
      if (b[j + 1][0] > 0)
        b[j][0] = b[j + 1][0] - 1, b[j][1] = b[j + 1][1];
      else
        b[j][1] = b[j + 1][1] + 1, b[j][0] = b[j + 1][0];
    }
    j--;
  }
  long long ans = 0;
  for (long long i = 1; i <= n; i++) {
    if (s1[i - 1] == '(') {
      if (a[i - 1][1] > 0)
        a[i - 1][1]--;
      else
        a[i - 1][0]++;
    } else
      a[i - 1][1]++;
    if (a[i - 1][0] == 0 && b[i + 1][1] == 0 && a[i - 1][1] == b[i + 1][0])
      ans++;
  }
  cout << ans << "\n";
}

### 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 long a[1000100][2], b[1000100][2];
int32_t main() {
  long long n;
  cin >> n;
  string s1;
  cin >> s1;
  stack<char> s;
  long long j = 1;
  for (char i : s1) {
    if (i == '(')
      a[j][1] = a[j - 1][1] + 1, a[j][0] = a[j - 1][0];
    else {
      if (a[j - 1][1] > 0)
        a[j][1] = a[j - 1][1] - 1, a[j][0] = a[j - 1][0];
      else
        a[j][0] = a[j - 1][0] + 1, a[j][1] = a[j - 1][1];
    }
    j++;
  }
  j = n;
  string s2 = s1;
  reverse(s2.begin(), s2.end());
  for (char i : s2) {
    if (i == ')')
      b[j][0] = b[j + 1][0] + 1, b[j][1] = b[j + 1][1];
    else {
      if (b[j + 1][0] > 0)
        b[j][0] = b[j + 1][0] - 1, b[j][1] = b[j + 1][1];
      else
        b[j][1] = b[j + 1][1] + 1, b[j][0] = b[j + 1][0];
    }
    j--;
  }
  long long ans = 0;
  for (long long i = 1; i <= n; i++) {
    if (s1[i - 1] == '(') {
      if (a[i - 1][1] > 0)
        a[i - 1][1]--;
      else
        a[i - 1][0]++;
    } else
      a[i - 1][1]++;
    if (a[i - 1][0] == 0 && b[i + 1][1] == 0 && a[i - 1][1] == b[i + 1][0])
      ans++;
  }
  cout << ans << "\n";
}

```