#include <bits/stdc++.h>
using namespace std;
int n, num[1100000], st[4400000];
string s;
void build(int a, int b, int index) {
  if (a == b)
    st[index] = num[a];
  else {
    build(a, (a + b) / 2, 2 * index);
    build((a + b) / 2 + 1, b, 2 * index + 1);
    st[index] = min(st[2 * index], st[2 * index + 1]);
  }
}
int query(int a, int b, int index, int l, int r) {
  if (a > r or b < l)
    return 1e9;
  else if (l <= a and b <= r) {
    return st[index];
  } else {
    return min(query(a, (a + b) / 2, 2 * index, l, r),
               query((a + b) / 2 + 1, b, 2 * index + 1, l, r));
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  cin >> s;
  int curr = 0;
  for (int i = 0; i < n; i++) {
    if (s[i] == '(')
      curr++;
    else
      curr--;
    num[i] = curr;
  }
  build(0, n - 1, 1);
  int ans = 0;
  for (int i = 0; i < n; i++) {
    if (s[i] == '(') {
      int minim = query(0, n - 1, 1, i, n - 1);
      if (num[n - 1] == 2 and minim >= 2) ans++;
    } else {
      int minim = query(0, n - 1, 1, i, n - 1);
      if (num[n - 1] == -2 and minim >= -2) ans++;
    }
    if (num[i] < 0) break;
  }
  cout << ans << endl;
}

### Prompt

Develop a solution in CPP 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>
using namespace std;
int n, num[1100000], st[4400000];
string s;
void build(int a, int b, int index) {
  if (a == b)
    st[index] = num[a];
  else {
    build(a, (a + b) / 2, 2 * index);
    build((a + b) / 2 + 1, b, 2 * index + 1);
    st[index] = min(st[2 * index], st[2 * index + 1]);
  }
}
int query(int a, int b, int index, int l, int r) {
  if (a > r or b < l)
    return 1e9;
  else if (l <= a and b <= r) {
    return st[index];
  } else {
    return min(query(a, (a + b) / 2, 2 * index, l, r),
               query((a + b) / 2 + 1, b, 2 * index + 1, l, r));
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  cin >> s;
  int curr = 0;
  for (int i = 0; i < n; i++) {
    if (s[i] == '(')
      curr++;
    else
      curr--;
    num[i] = curr;
  }
  build(0, n - 1, 1);
  int ans = 0;
  for (int i = 0; i < n; i++) {
    if (s[i] == '(') {
      int minim = query(0, n - 1, 1, i, n - 1);
      if (num[n - 1] == 2 and minim >= 2) ans++;
    } else {
      int minim = query(0, n - 1, 1, i, n - 1);
      if (num[n - 1] == -2 and minim >= -2) ans++;
    }
    if (num[i] < 0) break;
  }
  cout << ans << endl;
}

```