#include <bits/stdc++.h>
#pragma GCC optimize("-O2")
using namespace std;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  std::cerr << name << " = " << arg1 << std::endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  std::cerr.write(names, comma - names) << " = " << arg1 << " |";
  __f(comma + 1, args...);
}
const long double PI = acos(-1.0);
int pref[1000005] = {0}, suf[1000005] = {0};
int valpref[1000005], valsuf[1000005];
void solve() {
  int n;
  cin >> n;
  string s;
  cin >> s;
  valpref[0] = 1, valsuf[n] = 1;
  for (int i = 0; i < n; i++) {
    if (s[i] == '(')
      pref[i + 1] = pref[i] + 1;
    else
      pref[i + 1] = pref[i] - 1;
    if (s[n - i - 1] == '(')
      suf[n - i - 1] = suf[n - i] - 1;
    else
      suf[n - i - 1] = suf[n - i] + 1;
    valpref[i + 1] = valpref[i] * (pref[i + 1] >= 0);
    valsuf[n - i - 1] = valsuf[n - i] * (suf[n - i - 1] >= 0);
  }
  int ans = 0;
  for (int i = 0; i < n; i++) {
    if (!valpref[i] || !valsuf[i + 1]) continue;
    if (s[i] == '(') {
      if (pref[i] - 1 == suf[i + 1] && pref[i] > 0) ans++;
    } else if (pref[i] + 1 == suf[i + 1])
      ans++;
  }
  cout << ans;
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  int t;
  t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

### Prompt

Your challenge is to write a Cpp solution to the following problem:

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 GCC optimize("-O2")
using namespace std;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  std::cerr << name << " = " << arg1 << std::endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  std::cerr.write(names, comma - names) << " = " << arg1 << " |";
  __f(comma + 1, args...);
}
const long double PI = acos(-1.0);
int pref[1000005] = {0}, suf[1000005] = {0};
int valpref[1000005], valsuf[1000005];
void solve() {
  int n;
  cin >> n;
  string s;
  cin >> s;
  valpref[0] = 1, valsuf[n] = 1;
  for (int i = 0; i < n; i++) {
    if (s[i] == '(')
      pref[i + 1] = pref[i] + 1;
    else
      pref[i + 1] = pref[i] - 1;
    if (s[n - i - 1] == '(')
      suf[n - i - 1] = suf[n - i] - 1;
    else
      suf[n - i - 1] = suf[n - i] + 1;
    valpref[i + 1] = valpref[i] * (pref[i + 1] >= 0);
    valsuf[n - i - 1] = valsuf[n - i] * (suf[n - i - 1] >= 0);
  }
  int ans = 0;
  for (int i = 0; i < n; i++) {
    if (!valpref[i] || !valsuf[i + 1]) continue;
    if (s[i] == '(') {
      if (pref[i] - 1 == suf[i + 1] && pref[i] > 0) ans++;
    } else if (pref[i] + 1 == suf[i + 1])
      ans++;
  }
  cout << ans;
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  int t;
  t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

```