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 namespace std;
int n;
long long A[100050];
vector<map<long long, long long> > V;
map<long long, long long> M;
map<long long, long long>::iterator it;
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
int main() {
  cin >> n;
  for (int i = 0; (i) < (n); i++) cin >> A[i];
  V.clear();
  V = vector<map<long long, long long> >(n, map<long long, long long>());
  V[0][A[0]] = 1;
  for (int i = 1; (i) <= (n - 1); i++) {
    for (it = V[i - 1].begin(); it != V[i - 1].end(); it++) {
      long long gPrev = it->first;
      long long cPrev = it->second;
      long long gNew = gcd(gPrev, A[i]);
      V[i][gNew] += cPrev;
    }
    V[i][A[i]] += 1;
  }
  M.clear();
  for (int i = 0; (i) < (n); i++) {
    for (it = V[i].begin(); it != V[i].end(); it++) {
      M[it->first] += it->second;
    }
  }
  int q;
  long long t;
  cin >> q;
  while (q--) {
    cin >> t;
    cout << M[t] << endl;
  }
  return 0;
}