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
#include <bits/stdc++.h>
using namespace std;
int main() {
  int n;
  cin >> n;
  vector<pair<char, int> > vc;
  string s;
  cin >> s;
  for (int i = 0; i < s.length(); i++) {
    vc.push_back({s[i], i});
  }
  stack<pair<char, int> > st;
  for (int i = 0; i < n; i++) {
    if (st.size() && s[i] == ')' && st.top().first == '(') {
      st.pop();
    } else {
      st.push({s[i], i});
    }
  }
  if (st.size() != 2) {
    cout << 0 << endl;
  } else {
    int ans = 0;
    int y = st.top().second;
    st.pop();
    int x = st.top().second;
    if (s[x] != s[y]) {
      cout << 0 << endl;
      return 0;
    }
    if (s[x] == '(') {
      for (int i = y; i < n; i++) {
        ans += (s[i] == '(');
      }
    } else {
      for (int i = x; i >= 0; i--) {
        ans += (s[i] == ')');
      }
    }
    cout << ans << endl;
  }
}
