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>
const long long MAX = 1e6 + 5;
const long long MOD = 1e9 + 7;
using namespace std;
int sum[MAX], minn[MAX];
int main() {
  ios::sync_with_stdio(false);
  int n;
  string s;
  cin >> n;
  cin >> s;
  sum[0] = 0;
  for (int i = 0; i < n; i++) {
    if (s[i] == '(')
      sum[i + 1] = sum[i] + 1;
    else if (s[i] == ')')
      sum[i + 1] = sum[i] - 1;
  }
  minn[n] = sum[n];
  for (int i = n - 1; i >= 0; i--) minn[i] = min(minn[i + 1], sum[i]);
  int res = 0;
  for (int i = 0; i < n; i++) {
    if (sum[i] < 0) break;
    int t = sum[i];
    if (s[i] == ')')
      t++;
    else if (s[i] == '(')
      t--;
    if (t < 0) continue;
    if (sum[n] - sum[i + 1] + t != 0) continue;
    if (minn[i + 1] - sum[i + 1] + t < 0) continue;
    res++;
  }
  cout << res << '\n';
  return 0;
}
