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;
char st[1000005];
int minq[1000005], minh[1000005];
int f1[1000005], f2[1000005];
int n;
int main() {
  scanf("%d\n", &n);
  scanf("%s", st + 1);
  minq[0] = 1e9;
  for (int i = 1; i <= n; ++i) {
    if (st[i] == '(')
      f1[i] = f1[i - 1] + 1, f2[i] = f2[i - 1];
    else
      f1[i] = f1[i - 1], f2[i] = f2[i - 1] + 1;
    minq[i] = min(minq[i - 1], f1[i] - f2[i]);
  }
  minh[n + 1] = 1e9;
  for (int i = n; i >= 1; --i) minh[i] = min(minh[i + 1], f1[i] - f2[i]);
  int ans = 0;
  if (f1[n] - f2[n] == 2)
    for (int i = 1; i <= n; ++i)
      if (st[i] == '(' && minq[i - 1] >= 0 && minh[i] >= 2) {
        ans++;
      }
  if (f1[n] - f2[n] == -2)
    for (int i = 1; i <= n; ++i)
      if (st[i] == ')' && minq[i - 1] >= 0 && minh[i] >= -2) {
        ans++;
      }
  cout << ans << endl;
}
