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;
int arr[100008];
map<int, long long> res;
map<int, int> curm;
map<int, int> nxtm;
int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &arr[i]);
  }
  for (int i = 1; i <= n; ++i) {
    nxtm.clear();
    for (auto it : curm) {
      nxtm[gcd(it.first, arr[i])] += it.second;
    }
    nxtm[arr[i]]++;
    swap(curm, nxtm);
    for (auto it : curm) {
      res[it.first] += it.second;
    }
  }
  int q;
  scanf("%d", &q);
  while (q--) {
    int x;
    scanf("%d", &x);
    printf("%I64d\n", res[x]);
  }
  return 0;
}