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 JAVA):

Solution:
import java.io.*;
import java.util.HashMap;
import java.util.StringTokenizer;

public class Code {
    static long st[][] = new long[17][100000];


    public static void main(String[] args) throws IOException {
        Scanner sc = new Scanner(System.in);
        PrintWriter out = new PrintWriter(System.out);
        int n = sc.nextInt();
        int [] a = new int[n];
        HashMap<Integer,Long> map = new HashMap<>();
        for (int i = 0; i < n; i++)
            a[i] = sc.nextInt();
        SparseTable s = new SparseTable(a);
        for(int i=0;i<n;i++) {
            int idx = i;
            while(true){
                int lo = idx+1, hi = n-1, ans = n;
                int g = s.query(i, idx);
                while(lo <= hi){
                    int mid = (lo + hi)/2;
                    if(s.query(i, mid) < g)
                    {
                        ans = mid;
                        hi = mid-1;
                    }
                    else lo = mid+1;
                }
                map.put(g, map.getOrDefault(g, 0L) + ans-idx);
                idx = ans;
                if(ans == n)
                    break;
            }
        }
        int q = sc.nextInt();
        while (q-->0) {
            int x=sc.nextInt();
            out.println(map.getOrDefault(x,0L));
        }
        out.flush();
        out.close();
    }


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

        int A[], log [], SpT[][];

        SparseTable(int[] A)
        {
            int n = A.length;	this.A = A;
            int k = (int)Math.floor(Math.log(n) / Math.log(2)) + 1;
            SpT = new int[n][k];

            for (int i = 0; i < n; i++)
                SpT[i][0] = A[i]; 					// RMQ of sub array starting at index i and of length 2^0=1

            //overall time complexity = O(n log n)
            for (int j = 1; (1<<j) <= n; j++)
                for (int i = 0; i + (1<<j) - 1 < n; i++)
                    SpT[i][j] = gcd(SpT[i+(1<<(j-1))][j-1], SpT[i][j-1]);

            log = new int[n+1];
            for (int i = 1, cur = 0; i < log.length; i++) {
                log[i] = 1<<cur+1 < i?++cur:cur;
            }
        }

        int query(int i, int j)
        {
            int k = log[j-i+1]; // 2^k <= (j-i+1)

            return gcd(SpT[i][k], SpT[j-(1<<k)+1][k]);
        }
    }

    static class Scanner {
        StringTokenizer st;
        BufferedReader br;

        public Scanner(InputStream s) {
            br = new BufferedReader(new InputStreamReader(s));
        }

        public Scanner(FileReader fileReader) {
            br = new BufferedReader(fileReader);
        }

        public String next() throws IOException {
            while (st == null || !st.hasMoreTokens())
                st = new StringTokenizer(br.readLine());
            return st.nextToken();
        }

        public int nextInt() throws IOException {
            return Integer.parseInt(next());
        }

        public long nextLong() throws IOException {
            return Long.parseLong(next());
        }

        public String nextLine() throws IOException {
            return br.readLine();
        }

        public boolean ready() throws IOException {
            return br.ready();
        }
    }
}