#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 6;
int n;
int a[N], s[N], m[N], M[N];
int res;
int main() {
  n = ({
    long long first = 0;
    int c = getchar(), n = 0;
    for (; !isdigit(c); c = getchar()) n = (c == '-');
    for (; isdigit(c); c = getchar()) first = first * 10 + c - '0';
    n ? -first : first;
  });
  for (int i(1); i <= n; i++)
    a[i] = ({
             char c = getchar();
             while (c == ' ' || c == '\n') c = getchar();
             c;
           }) == '('
               ? 1
               : -1;
  for (int i(1); i <= n; i++) s[i] = s[i - 1] + a[i];
  m[n] = s[n];
  for (int i(n - 1); 1 <= i; i--) m[i] = min(s[i], m[i + 1]);
  M[1] = s[1];
  for (int i(2); i <= n; i++) M[i] = min(s[i], M[i - 1]);
  for (int i(1); i <= n; i++)
    if (M[i - 1] >= 0) {
      if (a[i] == 1 && s[n] == 2)
        if (m[i] >= 2) res++;
      if (a[i] == -1 && s[n] == -2)
        if (m[i] >= -2) res++;
    }
  cout << res;
}

### 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;
const int N = 1e6 + 6;
int n;
int a[N], s[N], m[N], M[N];
int res;
int main() {
  n = ({
    long long first = 0;
    int c = getchar(), n = 0;
    for (; !isdigit(c); c = getchar()) n = (c == '-');
    for (; isdigit(c); c = getchar()) first = first * 10 + c - '0';
    n ? -first : first;
  });
  for (int i(1); i <= n; i++)
    a[i] = ({
             char c = getchar();
             while (c == ' ' || c == '\n') c = getchar();
             c;
           }) == '('
               ? 1
               : -1;
  for (int i(1); i <= n; i++) s[i] = s[i - 1] + a[i];
  m[n] = s[n];
  for (int i(n - 1); 1 <= i; i--) m[i] = min(s[i], m[i + 1]);
  M[1] = s[1];
  for (int i(2); i <= n; i++) M[i] = min(s[i], M[i - 1]);
  for (int i(1); i <= n; i++)
    if (M[i - 1] >= 0) {
      if (a[i] == 1 && s[n] == 2)
        if (m[i] >= 2) res++;
      if (a[i] == -1 && s[n] == -2)
        if (m[i] >= -2) res++;
    }
  cout << res;
}

```