#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e6 + 10;
char kkm[maxn];
int cnt[maxn];
int main() {
  int n;
  scanf("%d", &n);
  scanf("%s", kkm + 1);
  int x = 0;
  for (int i = 1; i <= n; i++) {
    if (kkm[i] == '(')
      x++;
    else
      x--;
    cnt[i] = x;
  }
  for (int i = n - 1; i >= 1; i--) cnt[i] = min(cnt[i + 1], cnt[i]);
  int ans = 0, b = 0;
  for (int i = 1; i <= n; i++) {
    if (kkm[i] == '(') {
      if (cnt[i] == 2 && x == 2) ans++;
      b++;
    } else {
      if (cnt[i] == -2 && x == -2) ans++;
      b--;
    }
    if (b < 0) break;
  }
  printf("%d\n", ans);
  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>
using namespace std;
const int maxn = 1e6 + 10;
char kkm[maxn];
int cnt[maxn];
int main() {
  int n;
  scanf("%d", &n);
  scanf("%s", kkm + 1);
  int x = 0;
  for (int i = 1; i <= n; i++) {
    if (kkm[i] == '(')
      x++;
    else
      x--;
    cnt[i] = x;
  }
  for (int i = n - 1; i >= 1; i--) cnt[i] = min(cnt[i + 1], cnt[i]);
  int ans = 0, b = 0;
  for (int i = 1; i <= n; i++) {
    if (kkm[i] == '(') {
      if (cnt[i] == 2 && x == 2) ans++;
      b++;
    } else {
      if (cnt[i] == -2 && x == -2) ans++;
      b--;
    }
    if (b < 0) break;
  }
  printf("%d\n", ans);
  return 0;
}

```