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.*;
import java.io.*;

class Main
{

	

	private static void exhaustiveSearch(int[] a, boolean[] memo, int currentIndex, int sum) {

		memo[sum] = true;

		if (currentIndex >= a.length) 
			return;

		exhaustiveSearch(a, memo, currentIndex + 1, sum);
		exhaustiveSearch(a, memo, currentIndex + 1, sum + a[currentIndex]);
	}


	public static void main(String[] args) {
		try(Scanner reader = new Scanner(new InputStreamReader(System.in)))
		{

			boolean[] memo = new boolean[2001];
			for (int i = 0; i < memo.length; i++) 
				memo[i] = false;

			int n = reader.nextInt();
			reader.nextLine();
			int[] a = new int[n];

			for (int i = 0; i < a.length; i++) {
				a[i] = reader.nextInt();
			}

			Main.exhaustiveSearch(a, memo, 0, 0);

			reader.nextLine();
			
			int q = reader.nextInt();

			for (int i = 0; i < q; i++) 
				System.out.println(memo[reader.nextInt()] ? "yes" : "no");
			


		}catch (Exception e) {
			e.printStackTrace();
		}
		
	}
}