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;

class Program
{
	static int n;
	static int[] q;

	static void Main(string[] args)
	{
		n = int.Parse(Console.ReadLine());
		string[] sq = Console.ReadLine().Split(' ');

		q = new int[n];
		for (int i = 0; i < n; i++)
		{
			q[i] = int.Parse(sq[i]);
		}

		int a = int.Parse(Console.ReadLine());
		string[] sm = Console.ReadLine().Split(' ');
		for (int i = 0; i < a; i++)
		{
			int m = int.Parse(sm[i]);

			Console.WriteLine(Dfs(0, 0, m) ? "yes": "no");
		}
	}

	static bool Dfs(int i, int sum, int m)
	{
		if (i == n)
		{
			if (sum == m)
			{
				return true;
			}
			else
			{
				return false;
			}
		}

		if (Dfs(i + 1, sum , m))
		{
			return true;
		}

		if (Dfs(i + 1, sum + q[i] , m))
		{
			return true;
		}
		return false;
	}
}