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.HashSet;
import java.util.Scanner;
public class Main{
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int i;
        int[] give = new int [n];
        for(i=0;i<n;i++){
            give[i] = sc.nextInt();
        }
        int m = sc.nextInt();
        int j;
        int[] num = new int [m];
        for(j=0;j<m;j++){
            num[j] = sc.nextInt();
        }
        int k,l,o;
        HashSet<Integer> hs = new HashSet<>();
        for(k=0;k<Math.pow(2,n);k++){
            int sum = 0;
            for(l=0;l<n;l++){
                if((1&k>>l)==1){
                    sum += give[l];
                }
            }
            hs.add(sum);
        }
        for(o=0;o<m;o++){
            if(hs.contains(num[o])){
                System.out.println("yes");
            }else{
                System.out.println("no");
            }
        }
    }
}