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;
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long n;
  cin >> n;
  string s;
  cin >> s;
  vector<long long> pref(n, 0);
  vector<bool> ok(n, false);
  for (long long i = 0; i < n; i++) {
    if (s[i] == ')') {
      pref[i]--;
    } else {
      pref[i]++;
    }
    if (i > 0) pref[i] += pref[i - 1];
    if (i > 0 && !ok[i - 1])
      ok[i] = false;
    else if (pref[i] < 0)
      ok[i] = false;
    else
      ok[i] = true;
  }
  vector<long long> minSuff(n, 0);
  for (long long i = n - 1; i >= 0; i--) {
    minSuff[i] = pref[i];
    if (i != n - 1) minSuff[i] = min(minSuff[i], minSuff[i + 1]);
  }
  long long tot = 0;
  for (long long i = 0; i < n; i++) {
    if (s[i] == '(') {
      if (pref.back() != 2) continue;
      if (pref[i] - 2 < 0) continue;
      if (i > 0 && !ok[i - 1]) continue;
      if (i != n - 1 && minSuff[i + 1] < 2) continue;
      tot++;
    } else {
      if (pref.back() != -2) continue;
      if (i > 0 && !ok[i - 1]) continue;
      if (i != n - 1 && minSuff[i + 1] < -2) continue;
      tot++;
    }
  }
  cout << tot << endl;
}
