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 N = 1e6 + 6;
int op[N], cp[N], os[N], cs[N];
bool pok[N], sok[N];
int main() {
  int n;
  string s;
  cin >> n >> s;
  for (int i = 0; i < n; i++) {
    op[i] = (i ? op[i - 1] : 0) + (s[i] == '(');
    cp[i] = (i ? cp[i - 1] : 0) + (s[i] == ')');
    pok[i] = (i ? pok[i - 1] : true) & (op[i] >= cp[i]);
  }
  sok[n] = true;
  for (int i = n - 1; i >= 0; i--) {
    os[i] = os[i + 1] + (s[i] == '(');
    cs[i] = cs[i + 1] + (s[i] == ')');
    sok[i] = sok[i + 1] & (cs[i] >= os[i]);
  }
  int ans = 0;
  for (int i = 0; i < n; i++) {
    int open_or_close = (s[i] == '(') - (s[i] == ')');
    op[i] -= open_or_close;
    cp[i] += open_or_close;
    int pok_past = pok[i];
    pok[i] = (i ? pok[i - 1] : true) & (op[i] >= cp[i]);
    if (pok[i] && sok[i + 1] && op[i] + os[i + 1] == cp[i] + cs[i + 1]) ans++;
    pok[i] = pok_past;
    op[i] += open_or_close;
    cp[i] -= open_or_close;
  }
  cout << ans;
}
