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() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n, i;
  string second;
  cin >> n;
  cin >> second;
  if (n % 2) {
    cout << 0;
    return 0;
  }
  int max_neg = 0, sum = 0, ans = 0;
  for (i = 0; i < n; i++) {
    sum += (second[i] == '(' ? 1 : -1);
    max_neg = min(max_neg, sum);
  }
  if (sum == 2) {
    if (max_neg == 0) {
      sum = 0;
      int max_ind = 0;
      for (i = 0; i < n; i++) {
        sum += (second[i] == '(' ? 1 : -1);
        if (sum <= 1) max_ind = i;
      }
      for (i = max_ind + 1; i < n; i++) ans += (second[i] == '(');
    }
  } else if (sum == -2) {
    if (max_neg >= -2) {
      sum = 0;
      for (i = 0; i < n; i++) {
        sum += (second[i] == '(' ? 1 : -1);
        if (second[i] == ')') ++ans;
        if (sum < 0) break;
      }
    }
  }
  cout << ans;
}
