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;
bool debug = 1;
int n, m, k;
int dx[4] = {0, 1, 0, -1}, dy[4] = {1, 0, -1, 0};
long long ln, lk, lm;
int ggcd(int x, int y) {
  int z = y;
  while (x % y != 0) {
    z = x % y;
    x = y;
    y = z;
  }
  return z;
}
unordered_map<int, long long> mp, tmp, ans;
int main() {
  scanf("%d", &n);
  for (int(i) = 1; (i) <= (int)(n); (i)++) {
    scanf("%d", &k);
    tmp.clear();
    for (auto p : mp) tmp[ggcd(p.first, k)] += p.second;
    tmp[k]++;
    for (auto p : tmp) ans[p.first] += p.second;
    mp = tmp;
  }
  scanf("%d", &m);
  while (m--) {
    scanf("%d", &k);
    printf("%lld\n", ans[k]);
  }
  return 0;
}