#include <bits/stdc++.h>
using namespace std;
int n, ans;
int prefSuf[1000010][2];
string s;
void getPS(int k, char cl, char op) {
  stack<char> st;
  bool ok = 1;
  for (int i = 1; i <= n; ++i) {
    if (!ok) {
      prefSuf[i + k * (n - 2 * i + 1)][k] = -1e9;
      continue;
    }
    if (st.empty()) {
      if (s[i] == cl) {
        ok = 0;
        prefSuf[i + k * (n - 2 * i + 1)][k] = -1e9;
        continue;
      } else {
        st.push(s[i]);
      }
    } else {
      if (st.top() == op and s[i] == cl) {
        st.pop();
      } else {
        st.push(s[i]);
      }
    }
    prefSuf[i + k * (n - 2 * i + 1)][k] = st.size();
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> n >> s;
  s = " " + s + " ";
  getPS(0, ')', '(');
  reverse(s.begin(), s.end());
  getPS(1, '(', ')');
  reverse(s.begin(), s.end());
  for (int i = 1; i <= n; ++i) {
    if (s[i] == '(') {
      ans += (prefSuf[i - 1][0] == (prefSuf[i + 1][1] + 1));
    } else {
      ans += (prefSuf[i + 1][1] == (prefSuf[i - 1][0] + 1));
    }
  }
  cout << ans;
}

### Prompt

Your challenge is to write 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;
int n, ans;
int prefSuf[1000010][2];
string s;
void getPS(int k, char cl, char op) {
  stack<char> st;
  bool ok = 1;
  for (int i = 1; i <= n; ++i) {
    if (!ok) {
      prefSuf[i + k * (n - 2 * i + 1)][k] = -1e9;
      continue;
    }
    if (st.empty()) {
      if (s[i] == cl) {
        ok = 0;
        prefSuf[i + k * (n - 2 * i + 1)][k] = -1e9;
        continue;
      } else {
        st.push(s[i]);
      }
    } else {
      if (st.top() == op and s[i] == cl) {
        st.pop();
      } else {
        st.push(s[i]);
      }
    }
    prefSuf[i + k * (n - 2 * i + 1)][k] = st.size();
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> n >> s;
  s = " " + s + " ";
  getPS(0, ')', '(');
  reverse(s.begin(), s.end());
  getPS(1, '(', ')');
  reverse(s.begin(), s.end());
  for (int i = 1; i <= n; ++i) {
    if (s[i] == '(') {
      ans += (prefSuf[i - 1][0] == (prefSuf[i + 1][1] + 1));
    } else {
      ans += (prefSuf[i + 1][1] == (prefSuf[i - 1][0] + 1));
    }
  }
  cout << ans;
}

```