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 {
	public static void main(String[] arg) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int[] A = new int[n];
		for(int i=0; i<n; i++){
			A[i]=sc.nextInt();
		}
		int q = sc.nextInt();
		int[] M = new int[q];
		for(int i=0; i<q; i++){
			M[i]=sc.nextInt();
		}
		
		for(int i=0; i<q; i++){
			if(solve(n, 0, M[i], A) == 1){
				System.out.println("yes");
			} else {
				System.out.println("no");
			}
		}
	}
	
	public static int solve(int n, int i, int m, int[] A){
		int res=0;
		if(m==0) return 1;
		if(i>=n) return 0;
		if(solve(n, i+1, m, A)==1){
			res = solve(n, i+1, m, A);
		}else if(solve(n, i+1, m-A[i], A)==1){
			res = solve(n, i+1, m-A[i], A);
		}
		return res;
	}

}