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 long long MOD = 1e9 + 7;
const int INF = 0x3f3f3f3f;
const int MAXN = 2e5 + 7;
long long n;
long long a[MAXN];
map<long long, long long> ans;
map<long long, long long> last;
map<long long, long long> now;
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
int main() {
  ios::sync_with_stdio(false), cin.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n; i++) {
    now[a[i]]++;
    for (auto it = last.begin(); it != last.end(); ++it)
      now[gcd((*it).first, a[i])] += (*it).second;
    for (auto it = now.begin(); it != now.end(); ++it)
      ans[(*it).first] += (*it).second;
    last = now;
    now.clear();
  }
  int q;
  cin >> q;
  while (q--) {
    int x;
    cin >> x;
    cout << ans[x] << endl;
  }
  return 0;
}