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;
const int maxn = (int)2e5 + 10;
const int mod = (int)1e9 + 9;
const int inf = (1 << 30) - 1;
int n;
string s;
void solve() {
  cin >> n >> s;
  int bal = 0, mi = 0;
  for (int i = 0; i < n; i++) {
    if (s[i] == '(') {
      bal++;
    } else if (bal == 0) {
      mi++;
    } else {
      bal--;
    }
  }
  int cnt = 0;
  if (bal == 2 && mi == 0) {
    int cur = 0;
    for (int i = n - 1; i >= 0; i--) {
      if (s[i] == ')') {
        cur++;
      } else {
        cnt++;
        if (cur == 0) break;
        cur--;
      }
    }
  } else if (mi == 2 && bal == 0) {
    int cur = 0;
    for (int i = 0; i < n; i++) {
      if (s[i] == '(') {
        cur++;
      } else {
        cnt++;
        if (cur == 0) break;
        cur--;
      }
    }
  }
  cout << cnt << "\n";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  for (int i = 1; i <= t; i++) {
    solve();
  }
  return 0;
}
