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: Time Limit Exceeded)

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

class Program
{
	static void Main(string[] args)
	{
		int n = int.Parse(Console.ReadLine().Split()[0]);
		int[] qs = Console.ReadLine().Split().Select(str => int.Parse(str)).ToArray();
		int q = int.Parse(Console.ReadLine().Split()[0]);
		int[] ms = Console.ReadLine().Split().Select(str => int.Parse(str)).ToArray();

		for (int mi = 0; mi < q; mi++)
		{
			bool miHit = false;
			
			for (int j = 0; j < Math.Pow(2, n); j++)
			{
				var jbits = new BitArray(BitConverter.GetBytes(j));
				var sum = 0;
				for (int k = jbits.Count - 1; k >= 0; k--)
				{
					if (jbits[k])
					{
						sum += qs[k];
						if (sum == ms[mi])
						{
							miHit = true;
							break;
						}
					}
				}
				if (miHit)
				{
					break;
				}
			}
			System.Console.WriteLine(miHit ? "yes" : "no");
		}
	}
}