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 comment(linker, "/stack:247474112")
#pragma GCC optimize("Ofast")
using namespace std;
string s;
int n;
vector<int> Diff;
void Input() {
  cin >> n >> s;
  Diff.resize(n, 0);
}
void Solve() {
  int open = 0, close = 0;
  for (auto c : s) {
    open += (c == '(');
    close += (c == ')');
  }
  if (abs(open - close) != 2) {
    cout << "0\n";
    return;
  }
  for (int i = 0; i < n; i++) {
    if (i == 0) {
      if (s[i] == '(')
        Diff[0] = 1;
      else
        Diff[0] = -1;
    } else if (s[i] == '(')
      Diff[i] = Diff[i - 1] + 1;
    else if (s[i] == ')')
      Diff[i] = Diff[i - 1] - 1;
    if (Diff[i] < -2) {
      cout << "0\n";
      return;
    }
  }
  if (open > close && *min_element(Diff.begin(), Diff.end()) < 0) {
    cout << "0\n";
    return;
  }
  int ans = 0;
  if (Diff[n - 1] < 0) {
    int threshold = 0;
    while (threshold < n && Diff[threshold] >= 0) threshold++;
    for (int i = 0; i < n; i++) {
      if (s[i] == '(') continue;
      ans += (i <= threshold);
    }
  } else if (Diff[n - 1] > 0) {
    int threshold = n - 1;
    while (threshold >= 0 && Diff[threshold] > +1) threshold--;
    for (int i = 0; i < n; i++) {
      if (s[i] == ')') continue;
      ans += (i > threshold);
    }
  }
  cout << ans;
}
int main(int argc, char* argv[]) {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  Input();
  Solve();
  return 0;
}
