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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.HashSet;
import java.util.Set;
import java.util.regex.Pattern;

public class Main {
    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        int n = ip(br);
        Set<Integer> hList = new HashSet<>();
        hList.add(0);
        for (int i : createListInt(br)) {
            for (int j : hList.toArray(new Integer[0])) {
                hList.add(i + j);
            }
        }
        ip(br);
        for (int i : createListInt(br)) {
            System.out.println(hList.contains(i) ? "yes" : "no");
        }

    }

    static boolean isNumber(String num) {
        return Pattern.compile("^\\-?[0-9]*\\.?[0-9]+$").matcher(num).find();
    }

    static int[] createListInt(BufferedReader br) throws IOException {
        String[] lines = br.readLine().split(" ");
        int n = lines.length;
        int[] list = new int[n];
        while (n-- > 0) {
            list[n] = Integer.parseInt(lines[n]);
        }
        return list;
    }

    static int ip(BufferedReader s) throws IOException {
        return Integer.parseInt(s.readLine());
    }
}