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 main() {
  long long i, j, n, cnt = 0, flag = 0, k = 0;
  string str;
  cin >> n >> str;
  for (i = 0; i < n; i++) {
    if (str[i] == '(') {
      cnt++;
      if (cnt == 2) j = i;
    } else
      cnt--;
    if ((cnt == -2) && (!k)) k = i;
    if (cnt < -2) break;
    if (cnt < 0) flag = 1;
  }
  if ((!cnt) || (flag && cnt == 2) || (i != n) || ((cnt != 2) && (cnt != -2))) {
    cout << "0\n";
    return 0;
  }
  if (cnt == 2) {
    cnt = 0;
    for (i = j; i < n; i++) {
      if (str[i] == '(') cnt++;
    }
  } else {
    cnt = 0;
    for (i = 0; i < k; i++) {
      if (str[i] == '(')
        cnt++;
      else
        cnt--;
      if (cnt == -1) break;
    }
    cnt = 0;
    k = i;
    for (i = 0; i <= k; i++) {
      if (str[i] == ')') cnt++;
    }
  }
  cout << cnt << endl;
  return 0;
}
