#include <bits/stdc++.h>
using namespace std;
int i, j, m, n;
int r() {
  char ch = getchar();
  int ans = 0, f = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    ans *= 10;
    ans += ch - '0';
    ch = getchar();
  }
  return ans * f;
}
int a[1000010];
int pre[1000012], suf[1000011];
int main() {
  n = r();
  char x;
  for (i = 1; i <= n; i++) {
    x = getchar();
    if (x == '(')
      a[i] = 1;
    else
      a[i] = -1;
  }
  int st = n, en = 1;
  for (i = 1; i <= n; i++) {
    pre[i] = pre[i - 1] + a[i];
    if (pre[i] == -1) {
      st = i;
      break;
    }
  }
  for (i = n; i >= 1; i--) {
    suf[i] = suf[i + 1] + a[i];
    if (suf[i] == 1) {
      en = i;
      break;
    }
  }
  int cnt = 0;
  for (i = en; i <= st; i++) {
    if (a[i] == 1)
      if (pre[i - 1] + suf[i + 1] - 1 == 0) cnt++;
    if (a[i] == -1)
      if (pre[i - 1] + suf[i + 1] + 1 == 0) cnt++;
  }
  cout << cnt;
  return 0;
}

### Prompt

Create a solution in cpp for 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 i, j, m, n;
int r() {
  char ch = getchar();
  int ans = 0, f = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    ans *= 10;
    ans += ch - '0';
    ch = getchar();
  }
  return ans * f;
}
int a[1000010];
int pre[1000012], suf[1000011];
int main() {
  n = r();
  char x;
  for (i = 1; i <= n; i++) {
    x = getchar();
    if (x == '(')
      a[i] = 1;
    else
      a[i] = -1;
  }
  int st = n, en = 1;
  for (i = 1; i <= n; i++) {
    pre[i] = pre[i - 1] + a[i];
    if (pre[i] == -1) {
      st = i;
      break;
    }
  }
  for (i = n; i >= 1; i--) {
    suf[i] = suf[i + 1] + a[i];
    if (suf[i] == 1) {
      en = i;
      break;
    }
  }
  int cnt = 0;
  for (i = en; i <= st; i++) {
    if (a[i] == 1)
      if (pre[i - 1] + suf[i + 1] - 1 == 0) cnt++;
    if (a[i] == -1)
      if (pre[i - 1] + suf[i + 1] + 1 == 0) cnt++;
  }
  cout << cnt;
  return 0;
}

```