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;
  
class Main {
    static int[] aArrays = new int[20];
    static int gIndex = 0;
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        for (int i = 0; i < n; i++) {
            aArrays[i] = sc.nextInt();
            gIndex++;
        }
        int q = sc.nextInt();
        int[] mArrays = new int[200];
        int index = 0;
        for (int i = 0; i < q; i++) {
            mArrays[i] = sc.nextInt();
            index++;
        }
          
        for (int i = 0; i < index; i++) {
            if (isMatch(0, 0, mArrays[i])) {
                System.out.println("yes");
            } else {
                System.out.println("no");
            }
        }
          
    }
      
    public static boolean isMatch(int j, int S, int T) {
            if (S == T) {
                return true;
            }
            for (int i = j; i < gIndex; i++) {
                if (isMatch(i + 1, S + aArrays[i], T)) {
                    return true;
                }
            }
            return false;
        }
}