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;
int n;
string s;
int d[1000001];
int Min[1000001];
int main() {
  cin >> n;
  cin >> s;
  s = '.' + s;
  for (int i = 1; i <= n; i++) {
    d[i] = d[i - 1];
    if (s[i] == '(')
      d[i]++;
    else
      d[i]--;
  }
  Min[n] = d[n];
  for (int i = n - 1; i >= 1; i--) {
    Min[i] = min(d[i], Min[i + 1]);
  }
  int cnt = 0;
  for (int i = 1; i <= n; i++) {
    if (s[i] == '(') {
      if (d[n] == 2 && Min[i] >= 2) {
        cnt++;
      }
    }
    if (s[i] == ')') {
      if (d[n] == -2 && Min[i] >= -2) cnt++;
    }
    if (d[i] < 0) break;
  }
  cout << cnt;
  return 0;
}
