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