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 long long inf = 1e9;
const long long N = 2e5 + 1;
const long long bits = 20;
const long double pi = 3.14159265358979323846;
const long long mod = 1e9 + 7;
const long long p = 29;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  std::cout << std::fixed << std::setprecision(20);
  ;
  int n;
  cin >> n;
  vector<int> dp1(n + 2), dp2(n + 2);
  vector<int> v1(n + 2, 1), v2(n + 2, 1);
  string s;
  cin >> s;
  v1[0] = 1;
  for (int i = 1; i <= n; i++) {
    if (s[i - 1] == '(') {
      dp1[i] = dp1[i - 1] + 1;
    } else {
      dp1[i] = dp1[i - 1] - 1;
    }
    if (dp1[i] < 0) {
      v1[i] = 0;
    }
    v1[i] &= v1[i - 1];
  }
  v2[n + 1] = 1;
  for (int i = n; i >= 1; i--) {
    if (s[i - 1] == '(') {
      dp2[i] += dp2[i + 1] + 1;
    } else {
      dp2[i] += dp2[i + 1] - 1;
    }
    if (dp2[i] > 0) {
      v2[i] = 0;
    }
    v2[i] &= v2[i + 1];
  }
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    if (v1[i - 1] && v2[i + 1]) {
      if (s[i - 1] == ')') {
        if (dp1[i - 1] + 1 == -dp2[i + 1]) {
          ans++;
        }
      } else {
        if (dp1[i - 1] - 1 == -dp2[i + 1]) {
          ans++;
        }
      }
    }
  }
  cout << ans << "\n";
}
