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() {
  int size;
  string str;
  cin >> size >> str;
  vector<int> pref(size), suf(size);
  int begIndex = 0, endIndex = size - 1;
  for (int i = 0; i < size; i++) {
    pref[i] = (str[i] == '(' ? 1 : -1) + (i == 0 ? 0 : pref[i - 1]);
    if (pref[i] < 0) endIndex = min(i, endIndex);
  }
  for (int i = size - 1; i >= 0; i--) {
    suf[i] = (str[i] == '(' ? 1 : -1) + (i == size - 1 ? 0 : suf[i + 1]);
    if (suf[i] > 0) begIndex = max(i, begIndex);
  }
  int ans = 0;
  for (int i = begIndex; i <= endIndex; i++) {
    int evalLeft = (i == 0 ? 0 : pref[i - 1]),
        evalRight = (i == size - 1 ? 0 : suf[i + 1]);
    if ((evalLeft + evalRight == -1 && str[i] == ')') ||
        (evalLeft + evalRight == 1 && str[i] == '('))
      ans++;
  }
  cout << ans << endl;
  return 0;
}
