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;
vector<int> A(100001);
const int MAXLG = 18;
unordered_map<int, long long> SAVE;
int __gcd(int x, int y) {
  int save;
  for (; y != 0;) {
    save = y;
    y = x % y;
    x = save;
  }
  return x;
}
int main() {
  int N, M, Q, i, j, k, l;
  scanf("%d", &N);
  for (i = 0; i < N; ++i) scanf("%d", &A[i]);
  map<int, int> divisors;
  map<int, int> ndivisors;
  for (i = N - 1; i >= 0; --i) {
    ndivisors.clear();
    for (auto &p : divisors) ndivisors[__gcd(p.first, A[i])] += p.second;
    ndivisors[A[i]]++;
    for (auto &p : ndivisors) SAVE[p.first] += p.second;
    swap(ndivisors, divisors);
  }
  scanf("%d", &Q);
  for (; Q; --Q) {
    int x;
    scanf("%d", &x);
    printf("%I64d\n", SAVE.count(x) ? SAVE[x] : 0LL);
  }
  return 0;
}