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 = 1e5 + 5;
map<int, long long> mp1, mp2, res;
map<int, long long>::iterator it;
int n;
int a[MAXN];
int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
int main() {
  ios::sync_with_stdio(false);
  while (cin >> n) {
    mp1.clear(), mp2.clear();
    res.clear();
    for (int i = 0; i < n; i++) cin >> a[i];
    for (int i = 0; i < n; i++) {
      mp2[a[i]] = 1;
      if (res.find(a[i]) == res.end())
        res[a[i]] = 1;
      else
        res[a[i]]++;
      for (it = mp1.begin(); it != mp1.end(); it++) {
        int tmp = gcd(it->first, a[i]);
        if (mp2.find(tmp) == res.end())
          mp2[tmp] = it->second;
        else
          mp2[tmp] += it->second;
        if (res.find(tmp) == res.end())
          res[tmp] = it->second;
        else
          res[tmp] += it->second;
      }
      mp1 = mp2;
      mp2.clear();
    }
    int q;
    cin >> q;
    int x;
    for (int i = 0; i < q; i++) {
      cin >> x;
      cout << res[x] << endl;
    }
  }
  return 0;
}