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() {
  int n;
  string s;
  cin >> n;
  cin >> s;
  if (n % 2 == 1) {
    cout << 0;
    return 0;
  }
  int sum = 0;
  int sb = 0;
  int sa = 0;
  int i, j;
  for (i = 0; i < n; i++)
    if (s[i] == ')')
      sum--;
    else
      sum++;
  int res[1000006];
  for (i = 0; i <= n; i++) res[i] = 0;
  int ming = 0;
  for (i = 0; i < n; i++) {
    if (ming < 0) break;
    if (s[i] == '(')
      sa = sum - sb - 1;
    else
      sa = sum - sb + 1;
    if (s[i] == '(') {
      if ((sb - 1 >= 0) && (ming >= 0) && (sa + sb - 1 == 0)) res[i]++;
      sb++;
      ming = min(ming, sb);
    } else {
      if ((sb + 1 >= 0) && (ming >= 0) && (sa + sb + 1 == 0)) res[i]++;
      sb--;
      ming = min(ming, sb);
    }
  }
  std::reverse(s.begin(), s.end());
  for (i = 0; i < n; i++)
    if (s[i] == ')')
      s[i] = '(';
    else
      s[i] = ')';
  int count = 0;
  sum = 0;
  sa = 0;
  sb = 0;
  for (i = 0; i < n; i++)
    if (s[i] == ')')
      sum--;
    else
      sum++;
  ming = 0;
  for (i = 0; i < n; i++) {
    if (ming < 0) break;
    if (s[i] == '(')
      sa = sum - sb - 1;
    else
      sa = sum - sb + 1;
    if (s[i] == '(') {
      if ((sb - 1 >= 0) && (ming >= 0) && (sa + sb - 1 == 0))
        if (res[n - 1 - i] == 1) count++;
      sb++;
      ming = min(ming, sb);
    } else {
      if ((sb + 1 >= 0) && (ming >= 0) && (sa + sb + 1 == 0))
        if (res[n - 1 - i]++) count++;
      sb--;
      ming = min(ming, sb);
    }
  }
  cout << count;
}
