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.*;

public class Main {
    static int n;
    static int[] A = new int [50];

    static boolean solve(int i, int m) {
        // mになる組み合わせが見つかった
        if (m == 0) return true;
        // 与えられた数列の長さnを越えてしまった
        if (i >= n) return false;

        // 次の整数へ進める。i番目の整数を選んだ場合と選ばなかった場合
        boolean res = solve(i + 1, m) || solve(i + 1, m - A[i]);
        return res;
    }
    public static void main(String[] args) throws Exception {
        int q, m;

        Scanner sc = new Scanner(System.in);
        // 数列の長さ
        n = Integer.parseInt(sc.next());
        // 数列
        for (int i = 0; i < n; i++) {
            A[i] = Integer.parseInt(sc.next());
        }
        // mの数
        q = Integer.parseInt(sc.next());
        // q回検証する
        for (int i = 0; i < q; i++) {
            m = Integer.parseInt(sc.next());
            if (solve(0, m)) System.out.println("yes");
            else System.out.println("no");
        }
    }
}