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;
int in() {
  int x;
  scanf("%d", &x);
  return x;
}
long long lin() {
  long long x;
  scanf("%lld", &x);
  return x;
}
int main() {
  int n;
  string s;
  cin >> n >> s;
  vector<int> a(n + 1);
  for (int i = 0; i < n; ++i) {
    a[i + 1] = a[i];
    if (s[i] == '(')
      a[i + 1]++;
    else
      a[i + 1]--;
  }
  if (abs(a[n]) != 2) {
    cout << 0 << endl;
    return 0;
  }
  vector<int> mil(n, 1e9), mir(n, 1e9);
  mil[0] = 0;
  mir[n - 1] = a[n];
  for (int i = 0; i < n - 1; ++i) {
    mil[i + 1] = min(mil[i], a[i + 1]);
  }
  for (int i = n - 1; i >= 1; --i) {
    mir[i - 1] = min(mir[i], a[i]);
  }
  int ans = 0;
  if (a[n] == -2) {
    for (int i = 0; i < n; ++i) {
      if (s[i] == '(') continue;
      if (mil[i] >= 0 && mir[i] >= -2) ans++;
    }
  } else {
    for (int i = 0; i < n; ++i) {
      if (s[i] == ')') continue;
      if (mil[i] >= 0 && mir[i] >= 2) ans++;
    }
  }
  cout << ans << endl;
  return 0;
}
