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;
int GCD(int a, int b) {
  if (b == 0) return a;
  return (a % b == 0 ? b : GCD(b, a % b));
}
long long POW(long long base, long long exp) {
  long long val;
  val = 1;
  while (exp > 0) {
    if (exp % 2 == 1) {
      val = (val * base) % 1000000007;
    }
    base = (base * base) % 1000000007;
    exp = exp / 2;
  }
  return val;
}
int a[100005];
map<int, long long> ans;
map<int, int> mp, temp;
pair<int, int> pp;
int main() {
  int n, i, q, x, gcd;
  scanf("%d", &n);
  for (i = 0; i < n; i++) {
    scanf("%d", &a[i]);
  }
  for (i = 0; i < n; i++) {
    temp.clear();
    for (map<int, int>::iterator it = mp.begin(); it != mp.end(); ++it) {
      gcd = GCD(a[i], it->first);
      if (temp.count(gcd) == 1) {
        temp[gcd] += it->second;
      } else {
        temp[gcd] = it->second;
      }
    }
    if (temp.count(a[i]) == 1) {
      temp[a[i]] += 1;
    } else {
      temp[a[i]] = 1;
    }
    mp.clear();
    for (map<int, int>::iterator it = temp.begin(); it != temp.end(); ++it) {
      pp.first = it->first;
      pp.second = it->second;
      if (ans.count(pp.first) == 1) {
        ans[pp.first] = ans[pp.first] + pp.second;
      } else {
        ans[pp.first] = pp.second;
      }
      mp.insert(pp);
    }
  }
  scanf("%d", &q);
  while (q--) {
    scanf("%d", &x);
    cout << ans[x] << endl;
  }
  return 0;
}