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;



public class Main {
	static boolean brute(int[] A,int i,int x){
		if(x==0){
			return true;
		}
		if(i==A.length){
			return false;
		}
		if(brute(A,i+1,x)) return true;
		if(brute(A,i+1,x-A[i])) return true;
		return false;
	}
	public static void main(String[] args){
		Scanner scan=new Scanner(System.in);
		int n,q;
		n=scan.nextInt();
		int[] A=new int[n];
		for(int i=0;i<n;i++) A[i]=scan.nextInt();
		q=scan.nextInt();

		for(int i=0;i<q;i++){
			if(brute(A,0,scan.nextInt()) ) System.out.println("yes");
			else System.out.println("no");
		}
		
		
		
		scan.close();
	}
}