Problem:
Exhaustive Search

Write a program which reads a sequence A of n elements and an integer M, and outputs "yes" if you can make M by adding elements in A, otherwise "no". You can use an element only once.

You are given the sequence A and q questions where each question contains Mi.

Input

In the first line n is given. In the second line, n integers are given. In the third line q is given. Then, in the fourth line, q integers (Mi) are given.

Output

For each question Mi, print yes or no.

Constraints

n ≤ 20

q ≤ 200

1 ≤ elements in A ≤ 2000

1 ≤ Mi ≤ 2000

Sample Input 1

5
1 5 7 10 21
8
2 4 17 8 22 21 100 35

Sample Output 1

no
no
yes
yes
yes
yes
no
no

Notes

You can solve this problem by a Burte Force approach. Suppose solve(p, t) is a function which checkes whether you can make t by selecting elements after p-th element (inclusive). Then you can recursively call the following functions:

solve(0, M)

solve(1, M-{sum created from elements before 1st element})

solve(2, M-{sum created from elements before 2nd element})

...

The recursive function has two choices: you selected p-th element and not. So, you can check solve(p+1, t-A[p]) and solve(p+1, t) in solve(p, t) to check the all combinations.

For example, the following figure shows that 8 can be made by A[0] + A[2].

Solution in JAVA (judge verdict: Accepted)

import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        int[] a = new int[n];
        for (int i = 0; i < n; i++) {
            a[i] = in.nextInt();
        }

        int q = in.nextInt();
        int[] m = new int[q];
        for (int i = 0; i < q; i++) {
            m[i] = in.nextInt();
        }

        for (int mi : m) {
            if (solve(0, mi, a)) {
                System.out.println("yes");
            } else {
                System.out.println("no");
            }
        }
    }

    private static boolean solve(int p, int mi, int[] a) {
        if (mi == 0) {
            return true;
        }

        if (p == a.length || mi < 0) {
            return false;
        }

        return solve(p+1, mi, a) || solve(p+1, mi-a[p], a);
    }
}