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 sums[1001000] = {0};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  int n;
  cin >> n;
  string s;
  cin >> s;
  for (int i = 0; i < (int)(n); i++) {
    if (s[i] == '(')
      sums[i + 1] = sums[i] + 1;
    else
      sums[i + 1] = sums[i] - 1;
    if (sums[i] < -2) {
      cout << 0 << endl;
      return 0;
    }
  }
  if (abs(sums[n]) != 2) {
    cout << 0 << endl;
    return 0;
  }
  int t = sums[n];
  int minV = 2;
  int ans = 0;
  if (t < 0) {
    for (int i = 0; i < (int)(n); i++) {
      if (s[i] == ')') ans++;
      if (sums[i + 1] < 0) break;
    }
  } else {
    for (int i = 0; i < (int)(n); i++) {
      minV = min(minV, sums[i + 1]);
    }
    if (minV < 0) {
      cout << 0 << endl;
      return 0;
    }
    minV = 2;
    for (int i = (int)n - 1; i >= 0; i--) {
      minV = min(minV, sums[i + 1]);
      if (minV < 2) break;
      if (s[i] == '(') ans++;
    }
  }
  cout << ans << endl;
}
