#include <bits/stdc++.h>
using namespace std;
const long double EPS = 1e-9;
const long long MOD = 1e9 + 7;
int a[1000100];
bool p[1000100];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout << fixed << setprecision(20);
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  int n;
  cin >> n;
  string s;
  cin >> s;
  int cnt = 0;
  bool possible = true;
  p[0] = true;
  for (int i = 0; i < n; i++) {
    if (s[i] == '(')
      ++cnt;
    else
      --cnt;
    if (cnt < 0) possible = false;
    p[i + 1] = possible;
    a[i] = cnt;
  }
  for (int i = n - 2; i >= 0; i--) a[i] = min(a[i], a[i + 1]);
  int res = 0;
  for (int i = 0; i < n; i++) {
    if (s[i] == '(' && cnt - 2 == 0 && a[i] - 2 >= 0 && p[i])
      ++res;
    else if (s[i] == ')' && cnt + 2 == 0 && a[i] + 2 >= 0 && p[i])
      ++res;
  }
  cout << res << '\n';
  return 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;
const long double EPS = 1e-9;
const long long MOD = 1e9 + 7;
int a[1000100];
bool p[1000100];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout << fixed << setprecision(20);
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  int n;
  cin >> n;
  string s;
  cin >> s;
  int cnt = 0;
  bool possible = true;
  p[0] = true;
  for (int i = 0; i < n; i++) {
    if (s[i] == '(')
      ++cnt;
    else
      --cnt;
    if (cnt < 0) possible = false;
    p[i + 1] = possible;
    a[i] = cnt;
  }
  for (int i = n - 2; i >= 0; i--) a[i] = min(a[i], a[i + 1]);
  int res = 0;
  for (int i = 0; i < n; i++) {
    if (s[i] == '(' && cnt - 2 == 0 && a[i] - 2 >= 0 && p[i])
      ++res;
    else if (s[i] == ')' && cnt + 2 == 0 && a[i] + 2 >= 0 && p[i])
      ++res;
  }
  cout << res << '\n';
  return 0;
}

```