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 double pi = 3.1415926535897932384626433;
const long long M = 998244353;
long long n, t, a[1000100], k;
string s;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n >> s;
  if (n % 2) return cout << 0, 0;
  for (long long i = 0; i < n; i++) {
    if (s[i] == '(')
      t++;
    else
      t--;
    a[i] = t;
  }
  if (t == 2) {
    for (long long i = 0; i < n; i++) {
      if (a[i] < 0) return cout << 0, 0;
      if (a[i] >= 2) {
        if (s[i] == '(') k++;
      } else
        k = 0;
    }
  }
  if (t == -2) {
    for (long long i = n - 1; i >= 0; i--) {
      if (a[i] < -2) return cout << 0, 0;
      if (a[i] < 0) k = 0;
      if (a[i] > -2)
        if (s[i] == ')') k++;
    }
  }
  cout << k;
}
