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: Wrong Answer)

import java.util.Scanner;
public class Main {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner sc = new Scanner(System.in);
		int n,q,i,j,k,flg;
		n = sc.nextInt();
		int [] A = new int[n];
		for(i=0;i<n;i++){
			A[i] = sc.nextInt();
		}
		q = sc.nextInt();
		int [] M = new int[q];
		for(j=0;j<q;j++){
			M[j] = sc.nextInt();
		}
		
		for(i=0;i<q;i++){
			flg =0;
			for(j=0;j<=n;j++){
				if(flg==1) break;
				for(k=j+1;k<n;k++){
					if(M[i] == A[j]+A[k]) {
						flg =1;
						break;
					}
				}
			}
			if(flg==1){
				System.out.println("yes");
			}else{
				System.out.println("no");
			}
		}
	}

}