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::sync_with_stdio(0);
  cin.tie(0);
  int n, c = 0;
  string s;
  cin >> n >> s;
  int a = 0;
  int d[n];
  for (int i = 0; i < n; ++i) {
    if (s[i] == '(')
      ++a;
    else
      --a;
    d[i] = a;
  }
  for (int i = n - 2; i >= 0; --i) d[i] = min(d[i + 1], d[i]);
  int b = 0;
  for (int i = 0; i < n; ++i) {
    if (s[i] == '(') {
      if (d[i] - 2 >= 0 && a == 2) ++c;
      ++b;
    } else {
      if (d[i] + 2 >= 0 && a == -2) ++c;
      --b;
    }
    if (b < 0) break;
  }
  cout << c;
}
