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 C# (judge verdict: Accepted)


using System;
using System.Linq;
using System.Collections.Generic;


public class Test
{
	public static void Main( )
	{
		var n = int.Parse( Console.ReadLine().Trim() );
		var A = Console.ReadLine().Trim().Split().Select( x => int.Parse( x ) ).ToArray();
		var q = int.Parse( Console.ReadLine().Trim() );
		var M = Console.ReadLine().Trim().Split().Select( x => int.Parse( x ) );
		

		foreach( int i in M ) {
			if( solve( A, 0, i ) ) {
				Console.WriteLine( "yes" );
			} else {
				Console.WriteLine( "no" );
			}
		}
		
		return;
	}
	public static bool solve( int[] A, int p, int t )
	{
		if( t == 0 ) {
			return true;
		}

		if( p+1 <= A.Length ) {
			return solve( A, p+1, t-A[p] ) || solve( A, p+1, t );
		} else {
			return false;
		}
	}
}