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 st;
int call() {
  int s = 0, r = -1, neg = 0;
  for (int i = 0; st[i]; i++) {
    if (st[i] == '(')
      s++;
    else
      s--;
    if (s <= 1) r = i;
    if (s < 0) neg = 1;
  }
  int ans = 0;
  if (s == 2 && !neg) {
    s = 0;
    for (int i = 0; st[i]; i++) {
      if (st[i] == '(')
        s++;
      else
        s--;
      if (i > r) {
        if (st[i] == '(') {
          if (s - 2 >= 0) ans++;
        }
      }
    }
  }
  return ans;
}
int main() {
  cin >> n >> st;
  int ans = call();
  string y;
  for (int i = st.size() - 1; i >= 0; i--) {
    if (st[i] == ')')
      y += '(';
    else
      y += ')';
  }
  st = y;
  ans += call();
  cout << ans;
}
