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, a[100010], q;
map<int, long long> mp;
int c[100010], d[100010], tot;
int t1[100010], t2[100010];
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  c[0] = 0;
  d[0] = 0;
  mp[a[1]] = 1;
  c[++tot] = 1;
  d[tot] = a[1];
  for (int i = 2; i <= n; i++) {
    c[++tot] = i;
    d[tot] = a[i];
    for (int j = 1; j <= tot; j++) {
      d[j] = gcd(d[j], a[i]);
    }
    int tmp = 0;
    d[tot + 1] = 0;
    for (int j = 2; j <= tot + 1; j++) {
      if (d[j] != d[j - 1]) {
        tmp++;
        t1[tmp] = c[j - 1];
        t2[tmp] = d[j - 1];
      }
    }
    tot = tmp;
    for (int j = 1; j <= tot; j++) {
      c[j] = t1[j];
      d[j] = t2[j];
    }
    for (int j = 1; j <= tot; j++) {
      mp[d[j]] += (c[j] - c[j - 1]);
    }
  }
  scanf("%d", &q);
  while (q--) {
    int x;
    scanf("%d", &x);
    printf("%lld\n", mp[x]);
  }
  return 0;
}