Problem: Given a sequence of integers a1, ..., an and q queries x1, ..., xq on it. For each query xi you have to count the number of pairs (l, r) such that 1 ≤ l ≤ r ≤ n and gcd(al, al + 1, ..., ar) = xi.

<image> is a greatest common divisor of v1, v2, ..., vn, that is equal to a largest positive integer that divides all vi.

Input

The first line of the input contains integer n, (1 ≤ n ≤ 105), denoting the length of the sequence. The next line contains n space separated integers a1, ..., an, (1 ≤ ai ≤ 109).

The third line of the input contains integer q, (1 ≤ q ≤ 3 × 105), denoting the number of queries. Then follows q lines, each contain an integer xi, (1 ≤ xi ≤ 109).

Output

For each query print the result in a separate line.

Examples

Input

3
2 6 3
5
1
2
3
4
6


Output

1
2
2
0
1


Input

7
10 20 3 15 1000 60 16
10
1
2
3
4
5
6
10
20
60
1000


Output

14
0
2
2
2
0
2
2
1
1

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using std::cin;
using std::complex;
using std::cout;
using std::endl;
using std::fixed;
using std::greater;
using std::ios_base;
using std::lower_bound;
using std::make_pair;
using std::map;
using std::max;
using std::min;
using std::multiset;
using std::next_permutation;
using std::ostream;
using std::pair;
using std::prev_permutation;
using std::random_shuffle;
using std::reverse;
using std::set;
using std::setprecision;
using std::sort;
using std::string;
using std::swap;
using std::upper_bound;
using std::vector;
template <class C>
void mini(C& a4, C b4) {
  a4 = min(a4, b4);
}
template <class C>
void maxi(C& a4, C b4) {
  a4 = max(a4, b4);
}
template <class T1, class T2>
ostream& operator<<(ostream& out, pair<T1, T2> pair) {
  return out << "(" << pair.first << ", " << pair.second << ")";
}
map<long long, long long> res;
const long long N = 1e5 + 5;
long long nwd(long long a, long long b) {
  if (b == 0) {
    return a;
  }
  return nwd(b, a % b);
}
long long a[N];
int main() {
  ios_base::sync_with_stdio(0);
  cout << fixed << setprecision(10);
  double beg_clock = 1.0 * clock() / CLOCKS_PER_SEC;
  long long n;
  cin >> n;
  ;
  for (long long i = (1); i <= (n); ++i) {
    cin >> a[i];
  }
  vector<pair<long long, long long> > ints;
  ints.push_back(make_pair(1, a[1]));
  res[a[1]] = 1;
  for (long long i = (2); i <= (n); ++i) {
    vector<pair<long long, long long> > new_ints;
    ints.push_back(make_pair(i, a[i]));
    for (auto p : ints) {
      long long nw = nwd(p.second, a[i]);
      if (new_ints.empty() || nw != new_ints.back().second) {
        new_ints.push_back(make_pair(p.first, nwd(p.second, a[i])));
      }
    }
    res[new_ints.back().second] += i - new_ints.back().first + 1;
    for (long long j = 0; j < ((long long)(new_ints).size()) - 1; j++) {
      res[new_ints[j].second] -= new_ints[j].first - new_ints[j + 1].first;
    }
    ints = new_ints;
  }
  long long q;
  cin >> q;
  ;
  for (long long qq = (1); qq <= (q); ++qq) {
    long long x;
    cin >> x;
    ;
    cout << res[x] << "\n";
  }
  return 0;
}