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;
int n;
char s[1000010];
int a[1000010];
int sum[1000010];
void work1() {
  int cnt = 0;
  for (int i = 1; i <= n; i++) {
    if (a[i] == 1 && sum[i] >= 2) {
      cnt++;
    }
    if (sum[i] < 0) {
      cnt = 0;
      break;
    }
    if (sum[i] - 2 < 0) cnt = 0;
  }
  cout << cnt << endl;
}
void work2() {
  int cnt = 0;
  bool Jud = 1;
  for (int i = 1; i <= n; i++) {
    if (sum[i] >= -1 && a[i] == -1 && Jud) {
      cnt++;
    }
    if (sum[i] + 2 < 0) {
      cnt = 0;
      break;
    }
    if (sum[i] < 0) Jud = 0;
  }
  cout << cnt << endl;
}
int main() {
  scanf("%d\n", &n);
  gets(s);
  for (int i = 0; i < n; i++) a[i + 1] = (s[i] == '(' ? 1 : -1);
  for (int i = 1; i <= n; i++) sum[i] = sum[i - 1] + a[i];
  if (!(sum[n] == 2 || sum[n] == -2)) {
    printf("0");
    return 0;
  }
  if (sum[n] == 2)
    work1();
  else
    work2();
}
