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>
static const int MOD = 1000000007;
using ll = long long;
using u32 = unsigned;
using u64 = unsigned long long;
using namespace std;
template <class T>
constexpr T INF = ::numeric_limits<T>::max() / 32 * 15 + 208;
int main() {
  int n;
  cin >> n;
  string s;
  cin >> s;
  vector<int> s1(n + 1), s2(n + 1), m1(n + 1), m2(n + 1);
  for (int i = 0; i < n; ++i) {
    s1[i + 1] = s1[i] + (s[i] == '(' ? 1 : -1);
    m1[i + 1] = min(m1[i], s1[i + 1]);
  }
  for (int i = n - 1; i >= 0; --i) {
    int t = (s[i] == '(' ? 1 : -1);
    s2[i] = s2[i + 1] + t;
    m2[i] = min(m2[i + 1] + t, t);
  }
  int ans = 0;
  auto f = [&](pair<int, int> a, pair<int, int> b) -> pair<int, int> {
    return {a.first + b.first, min(a.second, b.second + a.first)};
  };
  for (int i = 0; i < n; ++i) {
    int t = (s[i] == '(' ? -1 : 1);
    auto p = f(f({s1[i], m1[i]}, {t, t}), {s2[i + 1], m2[i + 1]});
    if (p.first == 0 && p.second >= 0) ans++;
  }
  cout << ans << "\n";
  return 0;
}
