#include <bits/stdc++.h>
using namespace std;
int n, cnt;
char a[1000010];
int pre[1000010], post[1000010];
int sum[1000010], ans;
bool f = 0;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
    if (a[i] == '(')
      sum[i] = sum[i - 1] + 1;
    else
      sum[i] = sum[i - 1] - 1;
  }
  if (sum[n] != -2 && sum[n] != 2) {
    printf("0");
    return 0;
  }
  pre[0] = 0x3f3f3f3f;
  post[n + 1] = 0x3f3f3f3f;
  for (int i = 1; i <= n; ++i) pre[i] = min(pre[i - 1], sum[i]);
  for (int i = n; i != 0; --i) post[i] = min(post[i + 1], sum[i]);
  for (int i = 1; i <= n; ++i) {
    if (a[i] == '(') {
      if ((sum[n] == 2) && (post[i] >= 2) && (pre[i - 1] >= 0)) ++ans;
    } else {
      if ((sum[n] == -2) && (post[i] >= -2) && (pre[i - 1] >= 0)) ++ans;
    }
  }
  printf("%d", ans);
  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;
int n, cnt;
char a[1000010];
int pre[1000010], post[1000010];
int sum[1000010], ans;
bool f = 0;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
    if (a[i] == '(')
      sum[i] = sum[i - 1] + 1;
    else
      sum[i] = sum[i - 1] - 1;
  }
  if (sum[n] != -2 && sum[n] != 2) {
    printf("0");
    return 0;
  }
  pre[0] = 0x3f3f3f3f;
  post[n + 1] = 0x3f3f3f3f;
  for (int i = 1; i <= n; ++i) pre[i] = min(pre[i - 1], sum[i]);
  for (int i = n; i != 0; --i) post[i] = min(post[i + 1], sum[i]);
  for (int i = 1; i <= n; ++i) {
    if (a[i] == '(') {
      if ((sum[n] == 2) && (post[i] >= 2) && (pre[i - 1] >= 0)) ++ans;
    } else {
      if ((sum[n] == -2) && (post[i] >= -2) && (pre[i - 1] >= 0)) ++ans;
    }
  }
  printf("%d", ans);
  return 0;
}

```