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, q, c;
int a[100005];
map<int, long long> ans;
map<int, int> last;
map<int, int> now;
map<int, int>::iterator it;
int gcd(int a, int b) {
  if (!b) {
    return a;
  } else {
    return gcd(b, a % b);
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
  }
  ans.clear();
  for (int i = 0; i < n; i++) {
    swap(last, now);
    now.clear();
    for (it = last.begin(); it != last.end(); it++) {
      now[gcd(it->first, a[i])] += it->second;
    }
    now[a[i]]++;
    for (it = now.begin(); it != now.end(); it++) {
      ans[it->first] += it->second;
    }
  }
  scanf("%d", &q);
  for (int i = 0; i < q; i++) {
    scanf("%d", &c);
    printf("%lld\n", ans[c]);
  }
  return 0;
}