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;
using ll = long long;
int main() {
  int n;
  cin >> n;
  string s;
  cin >> s;
  int pref[n];
  pref[0] = ((s[0] == ')') ? -1 : 1);
  for (int i = 1; i < n; i++) {
    pref[i] = ((s[i] == ')') ? -1 : 1);
    pref[i] += pref[i - 1];
  }
  int msuf[n];
  msuf[n - 1] = pref[n - 1];
  for (int i = n - 2; i >= 0; i--) {
    msuf[i] = min(msuf[i + 1], pref[i]);
  }
  int mprev = 0;
  int count = 0;
  for (int i = 0; i < n; i++) {
    if (mprev < 0) break;
    if (msuf[i] + ((s[i] == ')') ? 2 : -2) == 0 &&
        pref[n - 1] + ((s[i] == ')') ? 2 : -2) == 0) {
      count++;
    }
    mprev = min(mprev, pref[i]);
  }
  cout << count << "\n";
}
