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 main() {
  long long(n);
  scanf("%lld", &(n));
  ;
  string(s);
  cin >> (s);
  ;
  vector<long long> dp(n, 0), dp2(n, 0);
  long long o = 0;
  for (long long i = 0; i < (n); i++) {
    if (s[i] == '(')
      dp[i] = ++o;
    else
      dp[i] = --o;
  }
  dp2[0] = dp[0];
  for (long long i = 1; i < (n); i++) dp2[i] = min(dp[i], dp2[i - 1]);
  for (long long i = n - 2; i >= (0); i--) dp[i] = min(dp[i + 1], dp[i]);
  long long res = 0;
  for (long long i = 0; i < (n); i++) {
    if (s[i] == '(') {
      if (dp[n - 1] == 2 && dp[i] >= 2 && (i == 0 || dp2[i - 1] >= 0)) res++;
    } else {
      if (dp[n - 1] == -2 && dp[i] >= -2 && (i == 0 || dp2[i - 1] >= 0)) res++;
    }
  }
  printf("%lld\n", res);
  return 0;
}
