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 maxk = 25, maxn = 100005;
int n, q;
int a[maxn], st[maxk][maxn], lg[maxn];
map<int, long long> ans;
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
int query(int l, int r) {
  int k = lg[r - l + 1];
  return gcd(st[k][l], st[k][r - (1 << k) + 1]);
}
int main() {
  scanf("%d", &n), lg[0] = -1;
  for (int i = 1; i <= n; i++)
    scanf("%d", &a[i]), st[0][i] = a[i], lg[i] = lg[i / 2] + 1;
  for (int i = 1; i <= 20; i++)
    for (int j = 1; j + (1 << (i - 1)) - 1 <= n; j++)
      st[i][j] = gcd(st[i - 1][j], st[i - 1][j + (1 << (i - 1))]);
  for (int i = 1; i <= n; i++)
    for (int j = i; j <= n; j++) {
      int v = query(i, j), L = j - 1, R = n + 1;
      while (L + 1 < R) {
        int mid = (L + R) >> 1;
        if (query(i, mid) != v)
          R = mid;
        else
          L = mid;
      }
      ans[v] += (L - j + 1), j = L;
    }
  scanf("%d", &q);
  for (int i = 1; i <= q; i++) {
    int x;
    scanf("%d", &x);
    if (ans.count(x))
      printf("%lld\n", ans[x]);
    else
      puts("0");
  }
  return 0;
}