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 res(string s, int l) {
  int sum = 0;
  for (int i = 0; i < l; i++) {
    if (s[i] == '(')
      sum++;
    else
      sum--;
    if (sum < -2) {
      return 0;
    }
  }
  return (sum);
}
int ressub0(string s, int l) {
  int sum = 0;
  for (int i = 0; i < l; i++) {
    if (s[i] == '(')
      sum++;
    else
      sum--;
    if (sum < 0) {
      return 0;
    }
  }
  return (sum);
}
int main() {
  string brackets;
  int brlength;
  cin >> brlength;
  cin >> brackets;
  int brsum = res(brackets, brlength);
  if (brsum == 2 && ressub0(brackets, brlength) == 2) {
    int openfree = 0;
    int counter = 0;
    for (int i = 0; i < brlength; i++) {
      if (brackets[i] == '(') {
        counter++;
        openfree++;
      } else {
        counter--;
      }
      if (counter == 2 && brackets[i] == '(') {
        openfree = 1;
      }
    }
    cout << openfree;
  } else if (brsum == -2) {
    int closedfree = 0;
    int counter = 0;
    for (int i = 0; i < brlength; i++) {
      if (brackets[i] == '(') {
        counter++;
      } else {
        counter--;
        closedfree++;
      }
      if (counter == -1) {
        cout << closedfree;
        break;
      }
    }
  } else {
    cout << 0;
  }
  return 0;
}
