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 = 2e6 + 5;
const int maxn = 1002;
long long pre[N];
long long mini[N];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long n;
  cin >> n;
  string second;
  cin >> second;
  if (second[0] == '(')
    pre[0] = 1;
  else
    pre[0] = -1;
  for (int i = 1; i < n; ++i) {
    if (second[i] == '(') {
      pre[i] = pre[i - 1] + 1;
    } else {
      pre[i] = pre[i - 1] - 1;
    }
  }
  mini[n - 1] = pre[n - 1];
  for (int i = n - 2; i >= 0; --i) {
    mini[i] = min(mini[i + 1], pre[i]);
  }
  long long c = 0;
  for (int i = 0; i < n; ++i) {
    if (second[i] == '(') {
      long long val = mini[i];
      if (val < 2) {
        if (pre[i] < 0) break;
        continue;
      } else if (pre[n - 1] == 2) {
        c++;
      }
    } else {
      long long val = mini[i];
      if (val < -2) {
        if (pre[i] < 0) break;
        continue;
      } else if (pre[n - 1] == -2) {
        c++;
      }
    }
    if (pre[i] < 0) break;
  }
  cout << c;
  return 0;
}
