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;
const int maxn = 100010;
map<int, int> M[2];
map<int, long long> result;
int n;
int gcd(int a, int b) {
  int t = a % b;
  if (t == 0)
    return b;
  else
    return gcd(b, t);
}
void init() {
  int i, j, x, y, old, now;
  map<int, int>::iterator p;
  scanf("%d", &n);
  now = 0;
  scanf("%d", &x);
  M[now][x] = 1;
  result[x] = 1;
  for (i = 2; i <= n; ++i) {
    old = now;
    now = 1 - now;
    M[now].clear();
    scanf("%d", &x);
    for (p = M[old].begin(); p != M[old].end(); ++p) {
      y = gcd(p->first, x);
      M[now][y] += p->second;
    }
    M[now][x] += 1;
    for (p = M[now].begin(); p != M[now].end(); ++p)
      result[p->first] += p->second;
  }
}
void output() {
  int m, i, x;
  scanf("%d", &m);
  for (i = 1; i <= m; ++i) {
    scanf("%d", &x);
    printf("%I64d\n", result[x]);
  }
}
int main() {
  init();
  output();
  return 0;
}