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 Left[1000010];
int mini[1000010];
int main(void) {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  string S;
  cin >> S;
  for (int i = 0; i < S.length(); i++) {
    if (S[i] == '(')
      Left[i + 1] = Left[i] + 1;
    else
      Left[i + 1] = Left[i] - 1;
  }
  for (int i = 0; i < 1000010; i++) mini[i] = 2000000000;
  int ans = 0;
  if (Left[n] == 2) {
    for (int i = n; i >= 0; i--) mini[i] = min(mini[i + 1], Left[i]);
    for (int i = 0; i < n; i++) {
      if (i != 0 && Left[i - 1] < 0) break;
      if (S[i] == '(' && mini[i + 1] >= 2 && Left[n] == 2) ans++;
    }
  } else if (Left[n] == -2) {
    for (int i = n; i >= 0; i--) mini[i] = min(mini[i + 1], Left[i]);
    for (int i = 0; i < n - 1; i++) {
      if (i != 0 && Left[i - 1] < 0) break;
      if (S[i] == ')' && mini[i] >= -2) ans++;
    }
  }
  cout << ans << endl;
  return 0;
}
