#include <bits/stdc++.h>
using namespace std;
template <class A>
void pr(A a) {
  cout << a;
  cout << '\n';
}
template <class A, class B>
void pr(A a, B b) {
  cout << a << ' ';
  pr(b);
}
template <class A, class B, class C>
void pr(A a, B b, C c) {
  cout << a << ' ';
  pr(b, c);
}
template <class A, class B, class C, class D>
void pr(A a, B b, C c, D d) {
  cout << a << ' ';
  pr(b, c, d);
}
template <class A>
void PR(A a, long long n) {
  for (long long i = (long long)(0); i < (long long)(n); i++) {
    if (i) cout << ' ';
    cout << a[i];
  }
  cout << '\n';
}
long long check(long long n, long long m, long long x, long long y) {
  return x >= 0 && x < n && y >= 0 && y < m;
}
const long long MAX = 1e9 + 7, MAXL = 1LL << 61, dx[4] = {-1, 0, 1, 0},
                dy[4] = {0, 1, 0, -1};
void Main() {
  long long n;
  string s;
  cin >> n >> s;
  long long d[n + 1];
  memset(d, -1, sizeof(d));
  d[0] = 0;
  long long x = 0;
  for (long long i = (long long)(0); i < (long long)(n); i++) {
    if (s[i] == '(')
      x++;
    else
      x--;
    if (x < 0) break;
    d[i + 1] = x;
  }
  x = 0;
  long long ans = 0;
  for (long long i = (long long)(n)-1; i >= 0; i--) {
    if (s[i] == ')') {
      if (d[i] != -1 && d[i] == x - 1) ans++;
      x++;
    } else {
      if (d[i] != -1 && d[i] == x + 1) ans++;
      x--;
    }
    if (x < 0) break;
  }
  pr(ans);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  Main();
  return 0;
}

### Prompt

Please formulate 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;
template <class A>
void pr(A a) {
  cout << a;
  cout << '\n';
}
template <class A, class B>
void pr(A a, B b) {
  cout << a << ' ';
  pr(b);
}
template <class A, class B, class C>
void pr(A a, B b, C c) {
  cout << a << ' ';
  pr(b, c);
}
template <class A, class B, class C, class D>
void pr(A a, B b, C c, D d) {
  cout << a << ' ';
  pr(b, c, d);
}
template <class A>
void PR(A a, long long n) {
  for (long long i = (long long)(0); i < (long long)(n); i++) {
    if (i) cout << ' ';
    cout << a[i];
  }
  cout << '\n';
}
long long check(long long n, long long m, long long x, long long y) {
  return x >= 0 && x < n && y >= 0 && y < m;
}
const long long MAX = 1e9 + 7, MAXL = 1LL << 61, dx[4] = {-1, 0, 1, 0},
                dy[4] = {0, 1, 0, -1};
void Main() {
  long long n;
  string s;
  cin >> n >> s;
  long long d[n + 1];
  memset(d, -1, sizeof(d));
  d[0] = 0;
  long long x = 0;
  for (long long i = (long long)(0); i < (long long)(n); i++) {
    if (s[i] == '(')
      x++;
    else
      x--;
    if (x < 0) break;
    d[i + 1] = x;
  }
  x = 0;
  long long ans = 0;
  for (long long i = (long long)(n)-1; i >= 0; i--) {
    if (s[i] == ')') {
      if (d[i] != -1 && d[i] == x - 1) ans++;
      x++;
    } else {
      if (d[i] != -1 && d[i] == x + 1) ans++;
      x--;
    }
    if (x < 0) break;
  }
  pr(ans);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  Main();
  return 0;
}

```