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, f[1000005], c[1000005], d[1000005], res = 0;
string s;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  cin >> n >> s;
  s = " " + s;
  for (int i = 1; i <= n; i++) {
    f[i] = f[i - 1];
    if (s[i] == '(')
      ++f[i];
    else
      --f[i];
  }
  c[n + 1] = 1;
  for (int i = n; i > 0; i--) {
    if (!c[i + 1]) {
      c[i] = 0;
      continue;
    }
    if (f[i] < -2)
      c[i] = 0;
    else
      c[i] = 1;
  }
  d[n + 1] = 1;
  for (int i = n; i > 0; i--) {
    if (!d[i + 1]) {
      d[i] = 0;
      continue;
    }
    if (f[i] < 2)
      d[i] = 0;
    else
      d[i] = 1;
  }
  for (int i = 1; i <= n; i++) {
    if (s[i] == '(') {
      if (d[i] && f[n] == 2) ++res;
    } else if (c[i] && f[n] == -2)
      ++res;
    if (f[i] < 0) break;
  }
  cout << res;
  return 0;
}
