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>
int const INF = (int)2e9;
using namespace std;
vector<int> a, t;
void build(int v, int tl, int tr) {
  if (tl == tr)
    t[v] = a[tl];
  else {
    int tm = (tl + tr) / 2;
    build(v * 2, tl, tm);
    build(v * 2 + 1, tm + 1, tr);
    t[v] = min(t[v * 2], t[v * 2 + 1]);
  }
}
int min(int v, int tl, int tr, int l, int r) {
  if (l > r) return INF;
  if (l == tl && r == tr) return t[v];
  int tm = (tl + tr) / 2;
  return min(min(v * 2, tl, tm, l, min(r, tm)),
             min(v * 2 + 1, tm + 1, tr, max(l, tm + 1), r));
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  string s;
  cin >> n >> s;
  a.resize(n), t.resize(4 * n);
  a[0] = (s[0] == '(' ? 1 : -1);
  for (int i = 1; i < n; ++i) a[i] = a[i - 1] + (s[i] == '(' ? 1 : -1);
  build(1, 0, n - 1);
  int ans = 0;
  for (int i = 0; i < n; ++i) {
    if (s[i] == '(') {
      if (a[n - 1] - 2 == 0 && min(1, 0, n - 1, 0, i - 1) >= 0 &&
          min(1, 0, n - 1, i, n - 1) >= 2)
        ans++;
    } else {
      if (a[n - 1] + 2 == 0 && min(1, 0, n - 1, 0, i - 1) >= 0 &&
          min(1, 0, n - 1, i, n - 1) + 2 >= 0)
        ans++;
    }
  }
  cout << ans;
}
