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
import java.util.HashMap;
import java.util.Scanner;

public class D {
	Scanner sc = new Scanner(System.in);

	void run() {
		int n = sc.nextInt();
		HashMap<Integer, Long> res = new HashMap<Integer, Long>();
		HashMap<Integer, Long> pre = new HashMap<Integer, Long>();
		for (int i = 0; i < n; i++) {
			int num = sc.nextInt();
			HashMap<Integer, Long> now = new HashMap<Integer, Long>();
			now.put(num, 1L);
			for (int key : pre.keySet()) {
				int gcd = gcd(num, key);
				now.put(gcd, now.containsKey(gcd) ? now.get(gcd) + pre.get(key) : pre.get(key));
			}
			for (int key : now.keySet()) {
				res.put(key, res.containsKey(key) ? res.get(key) + now.get(key) : now.get(key));
			}
			pre = now;
		}

		StringBuilder ans = new StringBuilder();
		int q = sc.nextInt();
		while (q-- > 0) {
			int x = sc.nextInt();
			ans.append((res.containsKey(x) ? res.get(x) : 0) + "\n");
		}
		System.out.print(ans);
	}

	int gcd(int a, int b) {
		return a % b == 0 ? b : gcd(b, a % b);
	}

	public static void main(String[] args) {
		new D().run();
	}
}