#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e6 + 100;
char st[maxn];
int dl[maxn], dr[maxn];
int main() {
  int n;
  while (cin >> n && n) {
    scanf("%s", st + 1);
    dl[0] = dr[n + 1] = 0;
    int num = 0;
    for (int i = 1; i <= n; ++i) {
      if (st[i] == '(')
        num++;
      else
        num--;
      if (num < 0 || dl[i - 1] < 0)
        dl[i] = -1;
      else
        dl[i] = num;
    }
    num = 0;
    for (int i = n; i >= 1; --i) {
      if (st[i] == ')')
        num++;
      else
        num--;
      if (num < 0 || dr[i + 1] < 0)
        dr[i] = -1;
      else
        dr[i] = num;
    }
    int ans = 0;
    for (int i = 1; i <= n; ++i) {
      if (dl[i - 1] >= 0 && dr[i + 1] >= 0) {
        int t = dl[i - 1] - dr[i + 1];
        if (t == 1 && st[i] == '(')
          ans++;
        else if (t == -1 && st[i] == ')')
          ans++;
      }
    }
    cout << ans << endl;
  }
  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 + 100;
char st[maxn];
int dl[maxn], dr[maxn];
int main() {
  int n;
  while (cin >> n && n) {
    scanf("%s", st + 1);
    dl[0] = dr[n + 1] = 0;
    int num = 0;
    for (int i = 1; i <= n; ++i) {
      if (st[i] == '(')
        num++;
      else
        num--;
      if (num < 0 || dl[i - 1] < 0)
        dl[i] = -1;
      else
        dl[i] = num;
    }
    num = 0;
    for (int i = n; i >= 1; --i) {
      if (st[i] == ')')
        num++;
      else
        num--;
      if (num < 0 || dr[i + 1] < 0)
        dr[i] = -1;
      else
        dr[i] = num;
    }
    int ans = 0;
    for (int i = 1; i <= n; ++i) {
      if (dl[i - 1] >= 0 && dr[i + 1] >= 0) {
        int t = dl[i - 1] - dr[i + 1];
        if (t == 1 && st[i] == '(')
          ans++;
        else if (t == -1 && st[i] == ')')
          ans++;
      }
    }
    cout << ans << endl;
  }
  return 0;
}

```