#include <bits/stdc++.h>
using namespace std;
const int N = 2222222;
int n, i, pre[N], f[N], g[N], a[N], ans;
char s[N];
inline bool check(int x, int A, int B) {
  if (pre[n] - A + B != 0) return 0;
  int o = min(f[x - 1], pre[x - 1] + B + g[x + 1]);
  if (o >= 0) return 1;
  return 0;
}
int main1() {
  scanf("%d%s", &n, s + 1);
  for (i = 1; i <= n; i++) a[i] = s[i] == '(' ? 1 : -1;
  for (i = 1; i <= n; i++) pre[i] = pre[i - 1] + a[i];
  for (i = 1; i <= n; i++) f[i] = min(f[i - 1], pre[i]);
  for (i = n; i; i--) {
    g[i] = min(0, a[i]);
    g[i] = min(g[i], g[i + 1] + a[i]);
  }
  for (i = 1; i <= n; i++)
    if (check(i, a[i], -a[i])) ans++;
  printf("%d", ans);
}
int main() {
  int n;
  string s;
  cin >> n >> s;
  vector<int> pref(n + 1), suf(n + 1);
  vector<bool> okp(n + 1), oks(n + 1);
  okp[0] = oks[n] = true;
  for (int i = 0; i < n; ++i) {
    pref[i + 1] = pref[i] + (s[i] == '(' ? +1 : -1);
    okp[i + 1] = okp[i] & (pref[i + 1] >= 0);
    suf[n - i - 1] = suf[n - i] + (s[n - i - 1] == ')' ? +1 : -1);
    oks[n - i - 1] = oks[n - i] & (suf[n - i - 1] >= 0);
  }
  int ans = 0;
  for (int i = 0; i < n; ++i) {
    if (!okp[i] || !oks[i + 1]) {
      continue;
    }
    if (s[i] == '(') {
      if (pref[i] - 1 - suf[i + 1] == 0) {
        ++ans;
      }
    } else if (s[i] == ')') {
      if (pref[i] + 1 - suf[i + 1] == 0) {
        ++ans;
      }
    }
  }
  cout << ans << endl;
  return 0;
}

### Prompt

In CPP, your task is to solve 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 N = 2222222;
int n, i, pre[N], f[N], g[N], a[N], ans;
char s[N];
inline bool check(int x, int A, int B) {
  if (pre[n] - A + B != 0) return 0;
  int o = min(f[x - 1], pre[x - 1] + B + g[x + 1]);
  if (o >= 0) return 1;
  return 0;
}
int main1() {
  scanf("%d%s", &n, s + 1);
  for (i = 1; i <= n; i++) a[i] = s[i] == '(' ? 1 : -1;
  for (i = 1; i <= n; i++) pre[i] = pre[i - 1] + a[i];
  for (i = 1; i <= n; i++) f[i] = min(f[i - 1], pre[i]);
  for (i = n; i; i--) {
    g[i] = min(0, a[i]);
    g[i] = min(g[i], g[i + 1] + a[i]);
  }
  for (i = 1; i <= n; i++)
    if (check(i, a[i], -a[i])) ans++;
  printf("%d", ans);
}
int main() {
  int n;
  string s;
  cin >> n >> s;
  vector<int> pref(n + 1), suf(n + 1);
  vector<bool> okp(n + 1), oks(n + 1);
  okp[0] = oks[n] = true;
  for (int i = 0; i < n; ++i) {
    pref[i + 1] = pref[i] + (s[i] == '(' ? +1 : -1);
    okp[i + 1] = okp[i] & (pref[i + 1] >= 0);
    suf[n - i - 1] = suf[n - i] + (s[n - i - 1] == ')' ? +1 : -1);
    oks[n - i - 1] = oks[n - i] & (suf[n - i - 1] >= 0);
  }
  int ans = 0;
  for (int i = 0; i < n; ++i) {
    if (!okp[i] || !oks[i + 1]) {
      continue;
    }
    if (s[i] == '(') {
      if (pref[i] - 1 - suf[i + 1] == 0) {
        ++ans;
      }
    } else if (s[i] == ')') {
      if (pref[i] + 1 - suf[i + 1] == 0) {
        ++ans;
      }
    }
  }
  cout << ans << endl;
  return 0;
}

```