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 maxn = 1e6 + 10;
int awsl[maxn];
int main() {
  int n;
  scanf("%d", &n);
  string s;
  cin >> s;
  vector<int> sum(n);
  vector<int> m(n);
  for (int i = 0; i < n; ++i) {
    if (s[i] == '(')
      awsl[i] = 1;
    else
      awsl[i] = -1;
  }
  sum[0] = awsl[0];
  for (int i = 1; i < n; ++i) {
    sum[i] = sum[i - 1] + awsl[i];
  }
  m.back() = sum.back();
  for (int i = n - 2; i >= 0; i--) {
    m[i] = min(sum[i], m[i + 1]);
  }
  int cnt = 0;
  for (int i = 0; i < n; ++i) {
    int v = 2 * awsl[i];
    if (m[i] - v >= 0 && sum.back() - v == 0) cnt++;
    if (sum[i] < 0) break;
  }
  cout << cnt << '\n';
}
