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;
bool perfect(string str) {
  int left = 0, right = 0;
  for (int i = 0; i < str.size(); i++) {
    if (str[i] == '(') left++;
    if (str[i] == ')') right++;
    if (left < right) return false;
  }
  if (left == right) return true;
  return false;
}
int main() {
  int n;
  cin >> n;
  string str(n, ' ');
  int left[n], right[n];
  left[0] = 0;
  right[0] = 0;
  for (int i = 0; i < n; i++) {
    cin >> str[i];
    if (i == 0)
      str[0] == '(' ? left[0]++ : right[0]++;
    else if (str[i] == '(') {
      left[i] = left[i - 1] + 1;
      right[i] = right[i - 1];
    } else {
      right[i] = right[i - 1] + 1;
      left[i] = left[i - 1];
    }
  }
  if (abs(left[n - 1] - right[n - 1]) != 2) {
    cout << 0;
    return 0;
  }
  for (int i = 0; i < n; i++) {
    if (left[i] - right[i] < 0) {
      str[i] = '(';
      if (perfect(str)) {
        cout << right[i];
        return 0;
      } else {
        cout << 0;
        return 0;
      }
    }
  }
  if (left[n - 1] - right[n - 1] == 2) {
    int i = n - 1;
    while (left[i] - right[i] >= 2) {
      i--;
    }
    cout << left[n - 1] - left[i];
    return 0;
  }
  cout << 0;
  return 0;
}
