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 all[1000010], n;
int minTo[1000010], total = 0;
string s;
int main() {
  cin >> n >> s;
  int temp = 0;
  for (int i = 0; i < n; ++i) {
    char c = s[i];
    if (c == '(')
      temp++;
    else
      temp--;
    all[i] = temp;
  }
  int _min = all[n - 1];
  for (int i = n - 1; i >= 0; --i) {
    _min = min(_min, all[i]);
    minTo[i] = _min;
  }
  _min = all[0];
  for (int i = 1; i < n - 1; ++i) {
    _min = min(_min, all[i - 1]);
    if (s[i] == ')') {
      if (minTo[i] + 2 < 0) continue;
      if (all[n - 1] + 2 != 0) continue;
      if (_min < 0) continue;
      if (all[0] != 1) continue;
      total++;
    } else if (s[i] == '(') {
      if (minTo[i] - 2 < 0) continue;
      if (all[n - 1] - 2 != 0) continue;
      if (_min < 0) continue;
      if (all[0] != 1) continue;
      total++;
    }
  }
  _min = min(_min, all[n - 2]);
  if (n > 1 && s[0] == ')' && minTo[0] + 2 >= 0 && all[n - 1] + 2 == 0) total++;
  if (n > 1 && s[n - 1] == '(' && _min >= 0 && all[0] == 1 &&
      all[n - 1] - 2 == 0)
    total++;
  cout << total << endl;
  return 0;
}
