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 oo = 1e9 + 7;
long long n, pref[1000005], sum, mnpref[1000005], mnsuff[1000005];
string s;
signed main() {
  ios_base::sync_with_stdio(0);
  cin >> n >> s;
  for (long long i = 0; i < n; i++)
    pref[i] = (!i ? 0 : pref[i - 1]) + (s[i] == '(' ? 1 : -1);
  if (abs(pref[n - 1]) != 2) {
    cout << 0;
    return 0;
  }
  for (long long i = 0; i < n; i++)
    mnpref[i] = min((!i ? oo : mnpref[i - 1]), pref[i]);
  mnsuff[n] = oo;
  for (long long i = n - 1; i >= 0; i--)
    mnsuff[i] = min(mnsuff[i + 1], pref[i]);
  if (pref[n - 1] == 2) {
    long long cnt = 0;
    for (long long i = 1; i < n; i++) {
      if (s[i] == ')') continue;
      cnt += (mnpref[i - 1] >= 0 && mnsuff[i] >= 2);
    }
    cout << cnt << "\n";
  } else {
    long long cnt = 0;
    for (long long i = 0; i < n; i++) {
      if (s[i] == '(') continue;
      cnt += ((!i || mnpref[i - 1] >= 0) && mnsuff[i] >= -2);
    }
    cout << cnt << "\n";
  }
}
