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.Collections.Generic;

namespace Project
{
	class MainClass
	{
		public static bool[,] dp;
		public static int n;
		public static int[] A;

		public static void Main(string[] args)
		{
			n = int.Parse(Console.ReadLine());
			A = new int[n];
			string[] s = Console.ReadLine().Split(' ');
			for (int i = 0; i < n; i++)
			{
				A[i] = int.Parse(s[i]);
			}
			int q = int.Parse(Console.ReadLine());
			int[] m = new int[q];
			s = Console.ReadLine().Split(' ');
			for (int i = 0; i < q; i++)
			{
				m[i] = int.Parse(s[i]);
			}
			foreach (int x in m)
			{
				dp = new bool[n + 1, x + 1];
				ope(x);
				Console.WriteLine(dp[0, x] ? "yes" : "no");
			}
		}

		public static void ope(int x)
		{
			for (int i = 0; i < n + 1; i++)
			{
				dp[i, 0] = true;
			}
			for (int i = n - 1; i >= 0; i--)
			{
				for (int j = 0; j < x + 1; j++)
				{
					dp[i, j] = dp[i + 1, j] || (j - A[i] < 0 ? false : dp[i + 1, j - A[i]]);
				}
			}
		}
	}
}