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 d[1000007];
int f[1000007];
int g[1000007];
int main() {
  int n;
  cin >> n;
  string s;
  cin >> s;
  for (int i = 0; i < s.size(); i++) {
    if (s[i] == '(')
      d[i + 1] = d[i] + 1;
    else
      d[i + 1] = d[i] - 1;
  }
  f[1] = d[1];
  for (int i = 2; i <= n; i++) f[i] = min(f[i - 1], d[i]);
  g[n] = d[n];
  for (int i = n - 1; i >= 1; i--) g[i] = min(g[i + 1], d[i]);
  int dem = 0;
  g[n + 1] = 10000;
  for (int i = 1; i <= n; i++) {
    if (s[i - 1] == '(') {
      if (d[i] - 2 >= 0 && f[i - 1] >= 0 && g[i + 1] - 2 >= 0 && d[n] - 2 == 0)
        dem++;
    } else if (d[i] + 2 >= 0 && f[i - 1] >= 0 && g[i + 1] + 2 >= 0 &&
               d[n] + 2 == 0)
      dem++;
  }
  cout << dem;
  return 0;
}
