#include <bits/stdc++.h>
#pragma optimize("Ofast")
using namespace std;
using ll = long long int;
using ull = unsigned long long int;
using dd = double;
using ldd = long double;
using si = short int;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
int main() {
  ios_base::sync_with_stdio(0);
  cout.tie(0);
  cin.tie(0);
  int n;
  cin >> n;
  string in;
  cin >> in;
  vector<int> pref(n);
  vector<int> suff(n);
  pref[0] = 2 * (in[0] == '(') - 1;
  suff.back() = 2 * (in.back() == ')') - 1;
  vector<bool> bp(n), bs(n);
  bp[0] = (pref[0] == -1);
  bs.back() = (suff.back() == -1);
  for (int i = 1; i < n; ++i) {
    pref[i] = pref[i - 1] + 2 * (in[i] == '(') - 1;
    bp[i] = ((pref[i] == -1) | bp[i - 1]);
    suff[n - 1 - i] = suff[n - i] + 2 * (in[n - i - 1] == ')') - 1;
    bs[n - 1 - i] = ((suff[n - 1 - i] == -1) | bs[n - i]);
  }
  int ans = 0;
  0;
  0;
  0;
  0;
  for (int i = 0; i < n; ++i) {
    if (i > 0 && bp[i - 1]) break;
    if (i < n - 1 && bs[i + 1]) continue;
    if ((i > 0 ? pref[i - 1] : 0) - (i < n - 1 ? suff[i + 1] : 0) == -1 &&
        in[i] != '(')
      ++ans;
    if ((i > 0 ? pref[i - 1] : 0) - (i < n - 1 ? suff[i + 1] : 0) == 1 &&
        in[i] != ')')
      ++ans;
  }
  cout << ans;
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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

### Response

```cpp
#include <bits/stdc++.h>
#pragma optimize("Ofast")
using namespace std;
using ll = long long int;
using ull = unsigned long long int;
using dd = double;
using ldd = long double;
using si = short int;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
int main() {
  ios_base::sync_with_stdio(0);
  cout.tie(0);
  cin.tie(0);
  int n;
  cin >> n;
  string in;
  cin >> in;
  vector<int> pref(n);
  vector<int> suff(n);
  pref[0] = 2 * (in[0] == '(') - 1;
  suff.back() = 2 * (in.back() == ')') - 1;
  vector<bool> bp(n), bs(n);
  bp[0] = (pref[0] == -1);
  bs.back() = (suff.back() == -1);
  for (int i = 1; i < n; ++i) {
    pref[i] = pref[i - 1] + 2 * (in[i] == '(') - 1;
    bp[i] = ((pref[i] == -1) | bp[i - 1]);
    suff[n - 1 - i] = suff[n - i] + 2 * (in[n - i - 1] == ')') - 1;
    bs[n - 1 - i] = ((suff[n - 1 - i] == -1) | bs[n - i]);
  }
  int ans = 0;
  0;
  0;
  0;
  0;
  for (int i = 0; i < n; ++i) {
    if (i > 0 && bp[i - 1]) break;
    if (i < n - 1 && bs[i + 1]) continue;
    if ((i > 0 ? pref[i - 1] : 0) - (i < n - 1 ? suff[i + 1] : 0) == -1 &&
        in[i] != '(')
      ++ans;
    if ((i > 0 ? pref[i - 1] : 0) - (i < n - 1 ? suff[i + 1] : 0) == 1 &&
        in[i] != ')')
      ++ans;
  }
  cout << ans;
  return 0;
}

```