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 = 1e6 + 7;
int n;
int a[N];
int b[N];
string s;
int mn[N];
int mx[N];
int f(char x) { return (x == '(' ? -1 : 1); }
int main() {
  cin >> n >> s;
  s = '0' + s;
  for (int i = 1; i <= n; i++) {
    a[i] = a[i - 1] + f(s[i]);
  }
  for (int i = n; i >= 1; i--) {
    b[i] = b[i + 1] + f(s[i]);
  }
  mx[0] = INT_MIN;
  for (int i = 1; i <= n; i++) {
    mx[i] = max(mx[i - 1], a[i]);
  }
  mn[n + 1] = INT_MAX;
  for (int i = n; i >= 1; i--) {
    mn[i] = min(mn[i + 1], b[i]);
  }
  int cnt = 0;
  for (int i = 1; i <= n; i++) {
    int x = f('(' + ')' - s[i]);
    if ((a[i - 1] + x + b[i + 1]) == 0) {
      if (mx[i - 1] <= 0 && (a[i - 1] + x) <= 0) {
        if (mn[i + 1] >= 0 && (b[i + 1] + x) >= 0) {
          cnt++;
        }
      }
    }
  }
  cout << cnt << "\n";
}
