#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 10;
int a[N], pre[N], post[N];
char s[N];
int n;
int main() {
  while (~scanf("%d", &n)) {
    scanf("%s", s + 1);
    int x = 0;
    memset(a, 0, sizeof(a));
    for (int i = 1; i <= n; i++) {
      if (s[i] == '(')
        x++;
      else
        x--;
      a[i] = x;
    }
    pre[0] = N, post[n] = N;
    for (int i = 1; i < n; i++) pre[i] = min(pre[i - 1], a[i]);
    for (int i = n - 1; i >= 0; i--) post[i] = min(post[i + 1], a[i]);
    int ans = 0;
    if (x != -2 && x != 2) {
      printf("0\n");
    } else {
      for (int i = 1; i <= n; i++) {
        if (s[i] == '(') {
          if (pre[i - 1] >= 0 && post[i] >= 2 && x == 2) ans++;
        } else if (pre[i - 1] >= 0 && post[i] >= -2 && x == -2)
          ans++;
      }
      printf("%d\n", ans);
    }
  }
}

### 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 N = 1e6 + 10;
int a[N], pre[N], post[N];
char s[N];
int n;
int main() {
  while (~scanf("%d", &n)) {
    scanf("%s", s + 1);
    int x = 0;
    memset(a, 0, sizeof(a));
    for (int i = 1; i <= n; i++) {
      if (s[i] == '(')
        x++;
      else
        x--;
      a[i] = x;
    }
    pre[0] = N, post[n] = N;
    for (int i = 1; i < n; i++) pre[i] = min(pre[i - 1], a[i]);
    for (int i = n - 1; i >= 0; i--) post[i] = min(post[i + 1], a[i]);
    int ans = 0;
    if (x != -2 && x != 2) {
      printf("0\n");
    } else {
      for (int i = 1; i <= n; i++) {
        if (s[i] == '(') {
          if (pre[i - 1] >= 0 && post[i] >= 2 && x == 2) ans++;
        } else if (pre[i - 1] >= 0 && post[i] >= -2 && x == -2)
          ans++;
      }
      printf("%d\n", ans);
    }
  }
}

```