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
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e6 + 10;
int sum[maxn], op[maxn];
int main() {
  int n, l = 0, r = 0, ans = 0, minn = 100;
  string s;
  cin >> n >> s;
  for (int i = 0; i < n; i++) {
    if (s[i] == '(') {
      op[i + 1] = 1;
      sum[i + 1] = sum[i] + 1;
      l++;
    } else {
      op[i + 1] = 0;
      sum[i + 1] = sum[i] - 1;
      r++;
    }
  }
  if (n & 1 == 1 || (abs(l - r) > 2) || (sum[n] == 0) ||
      (*min_element(sum + 1, sum + 1 + n) < -2) ||
      (*min_element(sum + 1, sum + 1 + n) < 0 && sum[n] != -2)) {
    printf("0\n");
    exit(0);
  }
  if (sum[n] == -2) {
    for (int i = 1; i <= n; i++) {
      if (op[i] == 0 && sum[i] < 0) {
        ans++;
        break;
      } else if (op[i] == 0 && sum[i] >= 0)
        ans++;
    }
  }
  if (sum[n] == 2) {
    for (int i = n; i >= 1; i--) {
      if (sum[i] >= 2)
        minn = i;
      else
        break;
    }
    minn = max(minn, 2);
    for (int i = minn; i <= n; i++) {
      if (op[i] == 1) ans++;
    }
  }
  cout << ans << endl;
  return 0;
}
