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;
long long int gcd(long long int a, long long int b) {
  return ((b == 0) ? a : gcd(b, a % b));
}
long long int a[100001];
map<long long int, long long int> mapp[100001], ans;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, TC;
  cin >> n;
  for (int i = 0; i < n; ++i) cin >> a[i];
  cin >> TC;
  mapp[0][a[0]] = 1LL;
  ans[a[0]] = 1;
  for (int i = 1; i <= n - 1; ++i) {
    mapp[i][a[i]] = 1;
    ans[a[i]]++;
    for (auto ptr = mapp[i - 1].begin(); ptr != mapp[i - 1].end(); ptr++) {
      long long int u = gcd(ptr->first, a[i]);
      mapp[i][u] += ptr->second, ans[u] += ptr->second;
    }
  }
  while (TC--) {
    int t;
    cin >> t;
    if (ans.find(t) != ans.end())
      cout << ans[t] << endl;
    else
      cout << 0 << endl;
  }
  return 0;
}