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;
const int N = 1e6 + 5;
const int inf = 1e9;
enum { prefix = 0, suffix };
int sum[2][N];
string s;
int n;
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0), cout.precision(15);
  cin >> n >> s;
  s = "$" + s;
  for (int i = (1); i <= ((n)); i += (+1)) {
    sum[prefix][i] = sum[prefix][i - 1] + (s[i] == '(' ? +1 : -1);
    if (sum[prefix][i] < 0) {
      sum[prefix][i] = -inf;
    }
  }
  for (int i = (n); i >= (1); i += (-1)) {
    sum[suffix][i] = sum[suffix][i + 1] + (s[i] == ')' ? +1 : -1);
    if (sum[suffix][i] < 0) {
      sum[suffix][i] = -inf;
    }
  }
  int ans = 0;
  for (int i = (1); i <= ((n)); i += (+1)) {
    int add = s[i] == '(' ? +1 : -1;
    if (sum[prefix][i - 1] - sum[suffix][i + 1] - add == 0) {
      ans++;
    }
  }
  cout << ans << '\n';
  return 0;
}
