#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 10;
int n;
char s[N], t[N];
int a[N];
int main() {
  scanf("%d%s", &n, t + 1);
  for (int i = 1; i <= n; ++i) {
    if (t[i] == '(')
      a[i] = a[i - 1] + 1;
    else
      a[i] = a[i - 1] - 1;
  }
  if (a[n] == 2) {
    for (int i = 1; i <= n; ++i) {
      s[i] = t[i];
    }
  }
  if (a[n] == -2) {
    for (int i = 1; i <= n; ++i) {
      if (t[n - i + 1] == '(')
        s[i] = ')';
      else
        s[i] = '(';
      if (s[i] == '(')
        a[i] = a[i - 1] + 1;
      else
        a[i] = a[i - 1] - 1;
    }
  }
  int res = 0;
  if (a[n] == 2) {
    int w = 0;
    bool ok = true;
    for (int i = n; i >= 1; --i) {
      if (a[i] < 0) {
        ok = false;
        break;
      }
      if (a[i] == 1 && w == 0) w = i;
    }
    if (ok) {
      for (int i = w + 1; i <= n; ++i) {
        if (s[i] == '(') res++;
      }
    }
  }
  printf("%d\n", res);
  return 0;
}

### Prompt

In cpp, your task is to solve 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 n;
char s[N], t[N];
int a[N];
int main() {
  scanf("%d%s", &n, t + 1);
  for (int i = 1; i <= n; ++i) {
    if (t[i] == '(')
      a[i] = a[i - 1] + 1;
    else
      a[i] = a[i - 1] - 1;
  }
  if (a[n] == 2) {
    for (int i = 1; i <= n; ++i) {
      s[i] = t[i];
    }
  }
  if (a[n] == -2) {
    for (int i = 1; i <= n; ++i) {
      if (t[n - i + 1] == '(')
        s[i] = ')';
      else
        s[i] = '(';
      if (s[i] == '(')
        a[i] = a[i - 1] + 1;
      else
        a[i] = a[i - 1] - 1;
    }
  }
  int res = 0;
  if (a[n] == 2) {
    int w = 0;
    bool ok = true;
    for (int i = n; i >= 1; --i) {
      if (a[i] < 0) {
        ok = false;
        break;
      }
      if (a[i] == 1 && w == 0) w = i;
    }
    if (ok) {
      for (int i = w + 1; i <= n; ++i) {
        if (s[i] == '(') res++;
      }
    }
  }
  printf("%d\n", res);
  return 0;
}

```