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;
long long mod = 1e9 + 7;
int n, sum[1000006], ans, mn[1000006];
string s;
int main() {
  cin >> n >> s;
  for (int i = 0; i < n; i++) {
    if (s[i] == '(')
      sum[i]++;
    else
      sum[i]--;
    sum[i] += sum[i - 1];
  }
  if (sum[n - 1] != 2 && sum[n - 1] != -2) {
    cout << 0;
    return 0;
  }
  if (sum[n - 1] == 2) {
    memset(sum, 0, sizeof sum);
    reverse(s.begin(), s.end());
    for (int i = 0; i < n; i++) {
      if (s[i] == '(')
        s[i] = ')';
      else
        s[i] = '(';
      if (s[i] == '(')
        sum[i]++;
      else
        sum[i]--;
      sum[i] += sum[i - 1];
    }
  }
  mn[n - 1] = sum[n - 1];
  for (int i = n - 2; i >= 0; i--) mn[i] = min(sum[i], mn[i + 1]);
  for (int i = 0; i < n; i++) {
    if (s[i] == ')' && mn[i + 1] == -2) {
      ans++;
    }
    if (sum[i] < 0) break;
  }
  cout << ans;
}
