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;
const int N = 1000006;
int n, s1[N], s2[N];
bool v1[N], v2[N];
char s[N];
int main() {
  cin >> n;
  scanf("%s", s + 1);
  for (int i = 1; i <= n; i++) s1[i] = s1[i - 1] + (s[i] == '(' ? 1 : -1);
  for (int i = 0; i <= n; i++)
    if (s1[i] >= 0)
      v1[i] = 1;
    else
      break;
  for (int i = n; i; i--) s2[i] = s2[i + 1] + (s[i] == ')' ? 1 : -1);
  for (int i = n + 1; i; i--)
    if (s2[i] >= 0)
      v2[i] = 1;
    else
      break;
  int ans = 0;
  for (int i = 1; i <= n; i++)
    if (v1[i - 1] && v2[i + 1]) {
      if (s[i] == '(') {
        if (s1[i - 1] > 0 && s1[i - 1] - 1 == s2[i + 1]) ans++;
      } else {
        if (s1[i - 1] + 1 == s2[i + 1]) ans++;
      }
    }
  cout << ans << endl;
  return 0;
}
