#include <bits/stdc++.h>
using namespace std;
const long long MAX = (long long)2e6 + 100;
long long mod = (long long)1e9 + 7, mod2 = (long long)1e9 + 9;
long long n, m, pref[MAX], suf[MAX];
void solve() {
  cin >> n;
  string s;
  cin >> s;
  long long bal = 0;
  for (int i = 1; i <= n; i++) {
    pref[i] = pref[i - 1];
    if (s[i - 1] == ')')
      pref[i]--;
    else
      pref[i]++;
    suf[i] = pref[i];
  }
  bal = pref[n];
  for (int i = 2; i <= n; i++) {
    pref[i] = min(pref[i], pref[i - 1]);
  }
  for (int i = n - 1; i > 0; i--) suf[i] = min(suf[i + 1], suf[i]);
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    long long dif = 0;
    if (s[i - 1] == ')')
      dif = 2;
    else
      dif = -2;
    if (pref[i - 1] >= 0 && suf[i] + dif >= 0 && bal + dif == 0) ans++;
  }
  cout << ans;
}
int main() {
  ios::sync_with_stdio(0);
  long long q;
  q = 1;
  while (q--) solve();
}

### Prompt

Please formulate 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>
using namespace std;
const long long MAX = (long long)2e6 + 100;
long long mod = (long long)1e9 + 7, mod2 = (long long)1e9 + 9;
long long n, m, pref[MAX], suf[MAX];
void solve() {
  cin >> n;
  string s;
  cin >> s;
  long long bal = 0;
  for (int i = 1; i <= n; i++) {
    pref[i] = pref[i - 1];
    if (s[i - 1] == ')')
      pref[i]--;
    else
      pref[i]++;
    suf[i] = pref[i];
  }
  bal = pref[n];
  for (int i = 2; i <= n; i++) {
    pref[i] = min(pref[i], pref[i - 1]);
  }
  for (int i = n - 1; i > 0; i--) suf[i] = min(suf[i + 1], suf[i]);
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    long long dif = 0;
    if (s[i - 1] == ')')
      dif = 2;
    else
      dif = -2;
    if (pref[i - 1] >= 0 && suf[i] + dif >= 0 && bal + dif == 0) ans++;
  }
  cout << ans;
}
int main() {
  ios::sync_with_stdio(0);
  long long q;
  q = 1;
  while (q--) solve();
}

```