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 main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> s;
  vector<int> pre(n + 3, 0), suf(n + 3, 0);
  int ans = 0;
  for (int i = 1, j = 0; i <= n; i++) {
    pre[i] = pre[i - 1];
    if (pre[i] == -1) continue;
    if (s[i - 1] == '(')
      j++;
    else
      j--;
    pre[i] = j;
  }
  for (int i = n, j = 0; i >= 1; i--) {
    suf[i] = suf[i + 1];
    if (suf[i] == 1) continue;
    if (s[i - 1] == '(')
      j++;
    else
      j--;
    suf[i] = j;
  }
  for (int i = 1; i <= n; i++) {
    int temp = 0;
    if (s[i - 1] == ')')
      temp++;
    else
      temp--;
    if (pre[i - 1] != -1 && suf[i + 1] != 1 &&
        pre[i - 1] + suf[i + 1] + temp == 0) {
      ans++;
    }
  }
  cout << ans << "\n";
  return 0;
}
