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 + 10;
int n;
char c[N];
int streak[N];
int lowestL[N];
int lowestR[N];
int main() {
  scanf("%d", &n);
  char newline;
  scanf("%c", &newline);
  for (int i = 1; i <= n; i++) scanf("%c", &c[i]);
  for (int i = 1; i <= n; i++)
    streak[i] = streak[i - 1] + (c[i] == '(' ? 1 : -1);
  for (int i = 1; i <= n; i++) lowestL[i] = min(lowestL[i - 1], streak[i]);
  lowestR[n + 1] = N;
  for (int i = n; i >= 1; i--) lowestR[i] = min(lowestR[i + 1], streak[i]);
  int ans = 0;
  for (int i = 1; i <= n; i++)
    if (lowestL[i - 1] >= 0) {
      if (c[i] == '(') {
        if (lowestR[i] >= 2 && streak[n] == 2) ans++;
      } else {
        if (lowestR[i] >= -2 && streak[n] == -2) ans++;
      }
    }
  cout << ans << endl;
  return 0;
}
