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 r[1000005];
int l[1000005];
bool f1[1000005];
bool f2[1000005];
int main() {
  ios_base::sync_with_stdio(false);
  int i, j, k, n, q, m, t, x, b;
  cin >> n;
  string s;
  cin >> s;
  int c = 0;
  bool flag = 0;
  for (i = n - 1; i >= 0; i--) {
    r[i] = c;
    f2[i] = flag;
    if (s[i] == ')') {
      c++;
    } else
      c--;
    if (c < 0) {
      flag = 1;
    }
  }
  flag = 0;
  int o = 0;
  for (i = 0; i < n; i++) {
    l[i] = o;
    f1[i] = flag;
    if (s[i] == '(') {
      o++;
    } else {
      o--;
    }
    if (o < 0) {
      flag = 1;
    }
  }
  int ans = 0;
  for (i = 0; i < n; i++) {
    if (f1[i] == 1 || f2[i] == 1) {
      continue;
    }
    if (s[i] == '(') {
      if (l[i] == r[i] + 1) {
        ans++;
      }
    } else {
      if (l[i] + 1 == r[i]) {
        ans++;
      }
    }
  }
  cout << ans;
  return 0;
}
