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;
import java.util.Arrays;
import java.util.List;

public class Main
{
	public static int A[] = new int[20];
	public static boolean rec(int s, int n, int now)
	{
		if(n == -1)
		{
			if(now == s)
			{
				return true;
			}
			return false;
		}
		return rec(s, n - 1, now + A[n]) | rec(s, n - 1, now);
	}
	public static void main(String args[])
	{
		Scanner sc = new Scanner(System.in);
		int n, q, m[] = new int[200];
		n = sc.nextInt();
		for(int i = 0; i < n; i++)
		{
			A[i] = sc.nextInt();
		}
		q = sc.nextInt();
		for(int i = 0; i < q; i++)
		{
			m[i] = sc.nextInt();
		}
		for(int i = 0; i < q; i++)
		{
			if(rec(m[i], n - 1, 0))
			{
				System.out.println("yes");
			}
			else
			{
				System.out.println("no");
			}
		}
		sc.close();
	}
}