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>
#pragma GCC optimize("Ofast")
using namespace std;
long long int gcd(long long int A, long long int B) {
  if (!B) return A;
  return gcd(B, A % B);
}
int main() {
  int n;
  string s;
  cin >> n >> s;
  vector<int> pbal(n + 2), sbal(n + 2);
  vector<bool> pref(n + 1, 1), suff(n + 1, 1);
  for (int i = 1; i <= n; i++) {
    if (s[i - 1] == '(')
      pbal[i]++;
    else
      pbal[i]--;
    pbal[i] += pbal[i - 1];
    if (pref[i - 1] && pbal[i] >= 0)
      pref[i] = 1;
    else
      pref[i] = 0;
  }
  for (int i = n; i > 0; i--) {
    if (s[i - 1] == ')')
      sbal[i]--;
    else
      sbal[i]++;
    sbal[i] += sbal[i + 1];
    if (suff[i + 1] && sbal[i] <= 0)
      suff[i] = 1;
    else
      suff[i] = 0;
  }
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    if (!pref[i - 1] || !suff[i + 1])
      continue;
    else {
      if (s[i - 1] == '(') {
        if (pbal[i - 1] > 0 && pbal[i - 1] - 1 + sbal[i + 1] == 0) ans++;
      } else {
        if (sbal[i + 1] < 0 && pbal[i - 1] + 1 + sbal[i + 1] == 0) ans++;
      }
    }
  }
  cout << ans;
}
