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;
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long N;
  cin >> N;
  string cad;
  cin >> cad;
  cad = "*" + cad;
  long long T[N + 2];
  long long TT[N + 2];
  long long acum = 0;
  for (long long i = 1; i <= N; i++) {
    if (cad[i] == '(')
      acum++;
    else
      acum--;
    T[i] = acum;
  }
  TT[N + 1] = 1e18;
  for (long long i = N; i >= 1; i--) {
    TT[i] = min(T[i], TT[i + 1]);
  }
  long long men = 1e18;
  long long cont = 0;
  for (long long i = 1; i <= N; i++) {
    if (cad[i] == '(' && men >= 0 && TT[i] - 2 >= 0 && acum - 2 == 0) cont++;
    if (cad[i] == ')' && men >= 0 && TT[i] + 2 >= 0 && acum + 2 == 0) cont++;
    men = min(men, T[i]);
  }
  cout << cont << '\n';
  return 0;
}
