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.IOException;
import java.io.InputStreamReader;

public class Main {
	private static int[] elements;
	public static void main(String[] args) {
		BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
	    
		try {
			int nico = Integer.parseInt(in.readLine());
			
			String[] se = in.readLine().split(" ");
			elements = new int[nico];
			for (int i = 0; i < nico; i++) {
				elements[i] = Integer.parseInt(se[i]);
			}
			
			int lim = Integer.parseInt(in.readLine());
			se = in.readLine().split(" ");
			
			int[] qs = new int[lim];
			for (int i = 0; i < nico; i++) {
				qs[i] = Integer.parseInt(se[i]);
			}
			
			for (int i = 0; i < lim; i++) {
				if (search(qs[i], 0) == true) System.out.println("yes");
				else System.out.println("no");
			}
		} catch (NumberFormatException e) {
			// TODO Auto-generated catch block
			e.printStackTrace();
		} catch (IOException e) {
			// TODO Auto-generated catch block
			e.printStackTrace();
		}
	}
	
	public static boolean search(int m, int i) {
		if (m == 0) return true;
		if (m < 0) return false;

		if (search(m - elements[i], i++) == false) return search(m, i++);
		return true;
	}
}