#include <bits/stdc++.h>
using namespace std;
const int MaxN = (int)1e6 + 9;
char str[MaxN + 9];
int csL[MaxN + 9];
int csR[MaxN + 9];
int dpL[MaxN + 9];
int dpR[MaxN + 9];
int f() {
  int n;
  cin >> n;
  cin >> (str + 1);
  for (int i = 1; i <= n; ++i) {
    csL[i] = csL[i - 1] + (str[i] == '(' ? 1 : -1);
    dpL[i] = min(dpL[i - 1], csL[i]);
  }
  for (int i = n; i >= 1; --i) {
    csR[i] = csR[i + 1] + (str[i] == ')' ? 1 : -1);
    dpR[i] = min(dpR[i + 1], csR[i]);
  }
  int res = 0;
  if (csL[n] == -2) {
    if (dpL[n] < -2) return 0;
    for (int i = 1; i <= n; ++i) {
      if (str[i] == ')') res++;
      if (csL[i] < 0) break;
    }
  } else if (csL[n] == 2) {
    if (dpR[1] < -2) return 0;
    for (int i = n; i >= 1; --i) {
      if (str[i] == '(') res++;
      if (csR[i] < 0) break;
    }
  }
  return res;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  cout << f();
  return 0;
}

### Prompt

Please create a solution in CPP 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 = (int)1e6 + 9;
char str[MaxN + 9];
int csL[MaxN + 9];
int csR[MaxN + 9];
int dpL[MaxN + 9];
int dpR[MaxN + 9];
int f() {
  int n;
  cin >> n;
  cin >> (str + 1);
  for (int i = 1; i <= n; ++i) {
    csL[i] = csL[i - 1] + (str[i] == '(' ? 1 : -1);
    dpL[i] = min(dpL[i - 1], csL[i]);
  }
  for (int i = n; i >= 1; --i) {
    csR[i] = csR[i + 1] + (str[i] == ')' ? 1 : -1);
    dpR[i] = min(dpR[i + 1], csR[i]);
  }
  int res = 0;
  if (csL[n] == -2) {
    if (dpL[n] < -2) return 0;
    for (int i = 1; i <= n; ++i) {
      if (str[i] == ')') res++;
      if (csL[i] < 0) break;
    }
  } else if (csL[n] == 2) {
    if (dpR[1] < -2) return 0;
    for (int i = n; i >= 1; --i) {
      if (str[i] == '(') res++;
      if (csR[i] < 0) break;
    }
  }
  return res;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  cout << f();
  return 0;
}

```