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;
long long n;
string s;
signed main() {
  ios::sync_with_stdio(0);
  cout.tie(0);
  cin.tie(0);
  cin >> n >> s;
  vector<long long> l(n, -1);
  vector<long long> r(n, -1);
  for (long long i = 0, bal = 0; i < n; ++i) {
    l[i] = bal;
    if (s[i] == '(')
      ++bal;
    else
      --bal;
    if (bal < 0) break;
  }
  for (long long i = n - 1, bal = 0; i >= 0; --i) {
    r[i] = bal;
    if (s[i] == ')')
      ++bal;
    else
      --bal;
    if (bal < 0) break;
  }
  long long ans = 0;
  for (long long i = 0; i < n; ++i) {
    if (l[i] != -1 && r[i] != -1) {
      if (s[i] == '(' && l[i] - 1 == r[i])
        ++ans;
      else if (s[i] == ')' && l[i] + 1 == r[i])
        ++ans;
    }
  }
  cout << ans << '\n';
  ;
}
