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;
import java.util.HashMap;
import java.util.Map;

public class Main{
    public static void main(String[] args){
        Scanner scan=new Scanner(System.in);
        int n=scan.nextInt();
        int[] a=new int[n];
        Map<Integer,Integer> mp=new HashMap<Integer,Integer>();
        for(int i=0;i<n;i++){
            a[i]=scan.nextInt();
        }
        for(int i=0;i<(1<<n);i++){
            int cnt=0;
            for(int j=0;j<n;j++){
                if((1&i>>j)==1) cnt+=a[j];
            }
            mp.put(cnt,1);
        }
        int q=scan.nextInt();
        while(q>0){
            int m=scan.nextInt();
            if(mp.containsKey(m)) System.out.println("yes");
            else System.out.println("no");
            q--;
        }
    }
}