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>
const int N = 1000001;
using namespace std;
int n;
int p[N];
char c[N];
bool good[N];
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n;
  good[0] = true;
  for (int i = 1; i <= n; i++) {
    cin >> c[i];
    p[i] = p[i - 1] + (c[i] == '(' ? 1 : -1);
    good[i] = good[i - 1];
    if (p[i] < 0) {
      good[i] = false;
    }
  }
  int res = 0;
  int gg = 1e9;
  for (int i = n; i >= 1; i--) {
    gg = min(gg, p[i]);
    if (good[i - 1] == false) {
      continue;
    }
    int dif = (c[i] == '(' ? -2 : 2);
    if (p[n] + dif == 0 && gg + dif >= 0) {
      res += 1;
    }
  }
  cout << res << "\n";
}
