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;
inline long long GCD(long long x, long long y) {
  long long r = x % y;
  while (r) {
    x = y;
    y = r;
    r = x % y;
  }
  return y;
}
const long long N = 1e6 + 69;
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(0);
  long long n;
  cin >> n;
  string s;
  cin >> s;
  long long a[n];
  long long add = 0;
  for (long long i = 0; i < n; i++) {
    if (s[i] == '(')
      a[i] = 1;
    else
      a[i] = -1;
    add += a[i];
  }
  if (n % 2 || add > 2 || add < -2 || add == 0 || add == 1 || add == -1) {
    cout << 0 << endl;
    return 0;
  }
  if (add == 2) {
    multiset<long long> lol;
    long long sum = 0;
    for (long long i = 0; i < n; i++) {
      sum += a[i];
      lol.insert(sum);
    }
    sum = 0;
    long long ans = 0;
    for (long long i = 0; i < n; i++) {
      sum += a[i];
      if (sum < 0) break;
      long long var = *lol.begin();
      var -= 2;
      if (var >= 0 && a[i] == 1) {
        ans++;
      }
      lol.erase(lol.find(sum));
    }
    cout << ans << '\n';
  } else {
    multiset<long long> lol;
    long long sum = 0;
    for (long long i = 0; i < n; i++) {
      sum += a[i];
      lol.insert(sum);
    }
    sum = 0;
    long long ans = 0;
    for (long long i = 0; i < n; i++) {
      sum += a[i];
      long long var = *lol.begin();
      var += 2;
      if (var >= 0 && a[i] == -1) {
        ans++;
      }
      lol.erase(lol.find(sum));
      if (sum < 0) break;
    }
    cout << ans << '\n';
  }
  return 0;
}
