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;
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  string str;
  cin >> n >> str;
  int psum[n + 1];
  psum[0] = 0;
  for (int i = 1; i <= n; i++) {
    psum[i] = psum[i - 1];
    if (str[i - 1] == '(')
      psum[i]++;
    else
      psum[i]--;
  }
  if (abs(psum[n]) != 2) {
    cout << "0\n";
    exit(0);
  }
  int left[n + 1];
  int right[n + 1];
  left[0] = 1e9;
  left[1] = psum[1];
  right[n] = psum[n];
  for (int i = 2; i <= n; i++) left[i] = min(left[i - 1], psum[i]);
  for (int i = n - 1; i >= 1; i--) right[i] = min(right[i + 1], psum[i]);
  int res = 0;
  for (int i = 1; i <= n; i++) {
    if (left[i - 1] >= 0 && right[i] >= -2 && psum[n] == -2 &&
        str[i - 1] == ')')
      res++;
    if (left[i - 1] >= 0 && right[i] >= 2 && psum[n] == 2 && str[i - 1] == '(')
      res++;
  }
  cout << res << '\n';
}
