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 n;
  cin >> n;
  string second;
  pair<char, int> c1, c2, q;
  cin >> second;
  stack<pair<char, int> > st;
  for (long i = 0; i < n; ++i) {
    if (!st.empty() && st.top().first != second[i] && second[i] == ')')
      st.pop();
    else
      st.push({second[i], i});
  }
  long kol = 0;
  if (st.size() == 2) {
    c1 = st.top();
    st.pop();
    c2 = st.top();
  } else {
    cout << 0;
    return 0;
  }
  if (c1.first == c2.first) {
    if (c1.first == '(') {
      for (long i = c1.second; i < n; ++i)
        if (second[i] == '(') ++kol;
    } else {
      for (long i = c2.second; i >= 0; --i)
        if (second[i] == ')') ++kol;
    }
    cout << kol;
  } else
    cout << 0;
  return 0;
}
