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 N = (int)1e5 + 7;
const int L = 20;
int n;
int lg[N];
int ggcd(int a, int b) {
  if (b == 0) return a;
  return ggcd(b, a % b);
}
int ret[N][L];
int get(int l, int r) {
  int k = lg[r - l + 1];
  return ggcd(ret[l][k], ret[r - (1 << k) + 1][k]);
}
map<int, long long> sol;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 2; i <= n; i++) lg[i] = 1 + lg[i / 2];
  for (int i = 1; i <= n; i++) cin >> ret[i][0];
  for (int k = 1; (1 << k) <= n; k++)
    for (int i = 1; i + (1 << k) - 1 <= n; i++)
      ret[i][k] = ggcd(ret[i][k - 1], ret[i + (1 << (k - 1))][k - 1]);
  for (int l = 1; l <= n; l++) {
    int r = l;
    while (r <= n) {
      int lo = r + 1, hi = n;
      int ans = r;
      while (lo <= hi) {
        int mid = (lo + hi) / 2;
        if (get(l, mid) == get(l, r)) {
          ans = mid;
          lo = mid + 1;
        } else
          hi = mid - 1;
      }
      sol[get(l, r)] += ans - r + 1;
      r = ans + 1;
    }
  }
  int q;
  cin >> q;
  while (q--) {
    int x;
    cin >> x;
    cout << sol[x] << "\n";
  }
}