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 MAX = 200000;
int n, a[MAX], q;
vector<pair<int, long long> > res, tmp;
map<int, long long> ans;
int gcd(int a, int b) {
  while (b > 0) {
    int r = a % b;
    a = b;
    b = r;
  }
  return a;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  res.push_back(pair<int, long long>(a[0], 1));
  ans.clear();
  ans[a[0]]++;
  for (int i = 1; i < n; i++) {
    tmp.clear();
    tmp = res;
    res.clear();
    tmp.push_back(pair<int, long long>(a[i], 1));
    for (int j = tmp.size() - 2; j >= 0; j--)
      tmp[j].first = gcd(tmp[j].first, a[i]);
    for (int j = 0; j < tmp.size(); j++)
      if (res.size() == 0 || tmp[j].first != res[res.size() - 1].first)
        res.push_back(tmp[j]);
      else
        res[res.size() - 1].second += tmp[j].second;
    for (int j = 0; j < res.size(); j++) ans[res[j].first] += res[j].second;
  }
  scanf("%d", &q);
  for (int i = 0; i < q; i++) {
    int x;
    scanf("%d", &x);
    printf("%lld\n", ans[x]);
  }
}