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;
int main() {
  int n, i;
  string s;
  while (cin >> n) {
    cin >> s;
    string rs = "";
    for (i = s.size() - 1; i >= 0; i--) rs += s[i];
    for (i = 0; i < n; i++)
      if (rs[i] == '(')
        rs[i] = ')';
      else
        rs[i] = '(';
    vector<int> pref(n + 1), suf(n + 1);
    vector<bool> okp(n + 1), oks(n + 1);
    okp[0] = oks[n] = true;
    for (i = 0; i < n; ++i) {
      pref[i + 1] = pref[i] + (s[i] == '(' ? 1 : -1);
      okp[i + 1] = okp[i] & (pref[i + 1] >= 0);
      suf[n - i - 1] = suf[n - i] + (rs[i] == '(' ? 1 : -1);
      oks[n - i - 1] = oks[n - i] & (suf[n - i - 1] >= 0);
    }
    int ans = 0;
    for (i = 0; i < n; ++i) {
      if (!okp[i] || !oks[i + 1]) continue;
      if (s[i] == '(') {
        if (pref[i] > 0 && pref[i] - 1 - suf[i + 1] == 0) ++ans;
      } else {
        if (pref[i] + 1 - suf[i + 1] == 0) ++ans;
      }
    }
    cout << ans << endl;
  }
  return 0;
}
