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.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.HashSet;
import java.util.Scanner;
import java.util.Set;

public class Main {

	public static void main(String[] args) {
		Scanner scan=new Scanner(new BufferedReader(new InputStreamReader(System.in)));
		int t=scan.nextInt();
		ArrayList<Integer> list=new ArrayList<Integer>();
		Set<Integer> set=new HashSet<Integer>();
		for (int i = 0; i < t; i++) {
			int x=scan.nextInt();
			list.add(x);
			set.add(x);

		}
		int c=scan.nextInt();
		for (int i = 0; i < c; i++) {
			int x=scan.nextInt();
			boolean f=true;
			for (int j = 0; j < list.size(); j++) {
				for (int k = j+1; k < list.size(); k++) {
					if((list.get(j)+list.get(k))==x){
						System.out.println("yes");
						f=false;
					}
				}
			}
			if(f){
				if(!set.add(x)){
					System.out.println("yes");
					f=false;
				}
			}

			if(f){
				System.out.println("no");
			}
		}

	}

}