#include <bits/stdc++.h>
using namespace std;
long long n, m, t, caseno, k;
struct info {
  int x, y;
  info() {}
  info(int x, int y) : x(x), y(y) {}
  bool operator<(const info &p) const {
    if (x == p.x) return y < p.y;
    return x < p.x;
  }
};
long long a[1000006];
char s[1000006];
int main() {
  long long i, j, f;
  scanf("%lld", &n);
  scanf("%s", &s);
  for (i = n; i >= 1; i--) s[i] = s[i - 1];
  long long res1 = 0;
  for (i = 1; i <= n; i++) {
    if (s[i] == '(')
      a[i] = 1;
    else
      a[i] = -1;
  }
  bool ok = 1;
  for (i = 1; i <= n; i++) {
    a[i] += a[i - 1];
    if (a[i] < 0) {
      ok = 0;
    }
  }
  if (ok) {
    if (a[n] == 2) {
      for (i = n; i >= 1; i--) {
        if (a[i] < 2) break;
        if (s[i] == '(') res1++;
      }
    }
  }
  long long res2 = 0;
  for (i = 1; i <= n; i++) {
    if (s[i] == '(')
      a[i] = 1;
    else
      a[i] = -1;
  }
  ok = 1;
  for (i = 1; i <= n; i++) {
    a[i] += a[i - 1];
    if (a[i] < -2) {
      ok = 0;
    }
  }
  if (ok) {
    if (a[n] == -2) {
      for (i = 1; i <= n; i++) {
        if (s[i] == ')') res2++;
        if (a[i] < 0) break;
      }
    }
  }
  long long res = res1 + res2;
  printf("%lld\n", res);
  return 0;
}

### Prompt

Generate a CPP solution 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;
long long n, m, t, caseno, k;
struct info {
  int x, y;
  info() {}
  info(int x, int y) : x(x), y(y) {}
  bool operator<(const info &p) const {
    if (x == p.x) return y < p.y;
    return x < p.x;
  }
};
long long a[1000006];
char s[1000006];
int main() {
  long long i, j, f;
  scanf("%lld", &n);
  scanf("%s", &s);
  for (i = n; i >= 1; i--) s[i] = s[i - 1];
  long long res1 = 0;
  for (i = 1; i <= n; i++) {
    if (s[i] == '(')
      a[i] = 1;
    else
      a[i] = -1;
  }
  bool ok = 1;
  for (i = 1; i <= n; i++) {
    a[i] += a[i - 1];
    if (a[i] < 0) {
      ok = 0;
    }
  }
  if (ok) {
    if (a[n] == 2) {
      for (i = n; i >= 1; i--) {
        if (a[i] < 2) break;
        if (s[i] == '(') res1++;
      }
    }
  }
  long long res2 = 0;
  for (i = 1; i <= n; i++) {
    if (s[i] == '(')
      a[i] = 1;
    else
      a[i] = -1;
  }
  ok = 1;
  for (i = 1; i <= n; i++) {
    a[i] += a[i - 1];
    if (a[i] < -2) {
      ok = 0;
    }
  }
  if (ok) {
    if (a[n] == -2) {
      for (i = 1; i <= n; i++) {
        if (s[i] == ')') res2++;
        if (a[i] < 0) break;
      }
    }
  }
  long long res = res1 + res2;
  printf("%lld\n", res);
  return 0;
}

```