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 INF = (int)1E9;
int gcd(int x, int y) { return y == 0 ? x : gcd(y, x % y); }
int N;
int a[100005];
int C, NC;
pair<int, int> cp[35], ncp[35];
int main() {
  cin >> N;
  map<int, long long> ans;
  for (int i = (0); i < (N); i++) {
    scanf("%d", a + i);
    NC = 0;
    ncp[NC++] = pair<int, int>(a[i], i);
    int g = a[i], pg = g;
    for (int j = (0); j < (C); j++) {
      g = gcd(g, cp[j].first);
      if (g != pg) {
        ncp[NC++] = pair<int, int>(g, cp[j].second);
        pg = g;
      }
    }
    for (int j = (0); j < (NC - 1); j++) {
      ans[ncp[j].first] += ncp[j].second - ncp[j + 1].second;
    }
    ans[ncp[NC - 1].first] += ncp[NC - 1].second + 1;
    memcpy(cp, ncp, sizeof(cp));
    C = NC;
  }
  int Q;
  cin >> Q;
  while (Q--) {
    int x;
    scanf("%d", &x);
    printf("%I64d\n", ans[x]);
  }
  return 0;
}