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
#include <bits/stdc++.h>
using namespace std;
const double eps = 1e-9;
const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const long long inf = (((long long)1) << 61) + 5;
const int N = 100005;
map<int, long long> mp, mp1, mp2;
map<int, long long>::iterator it;
int a[N];
int gcd(int x, int y) { return y == 0 ? x : gcd(y, x % y); }
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  mp[a[0]]++, mp1[a[0]]++;
  for (int i = 1; i < n; i++) {
    mp[a[i]]++;
    mp2[a[i]]++;
    for (it = mp1.begin(); it != mp1.end(); it++) {
      int x = gcd(it->first, a[i]);
      mp2[x] += it->second;
      mp[x] += it->second;
    }
    mp1.clear();
    for (it = mp2.begin(); it != mp2.end(); it++) {
      mp1[it->first] = it->second;
    }
    mp2.clear();
  }
  int q;
  scanf("%d", &q);
  while (q--) {
    int x;
    scanf("%d", &x);
    printf("%I64d\n", mp[x]);
  }
  return 0;
}
