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;
using static System.Console;
using System.Text;
using System.IO;

namespace AOJ
{
    class Program
    {
        static public long[] Sarray() { return ReadLine().Split().Select(long.Parse).ToArray(); }
        static public List<long> Slist() { return ReadLine().Split().Select(long.Parse).ToList(); }
        static void Main(string[] args)
        {
            var N = Sarray()[0];
            var Ai = Sarray();
            var Q = Sarray()[0];
            var Mi = Sarray();

            Func<long, long, bool> solve = null;
            solve = (fi, fm) =>
             {
                 if (fm == 0) return true;
                 if (N <= fi) return false;
                 return solve(fi + 1, fm) || solve(fi + 1, fm - Ai[fi]);
             };
            for (var i = 0; i < Q; ++i)
            {
                if (solve(0, Mi[i]))
                    WriteLine("yes");
                else
                    WriteLine("no");
            }
        }
    }
}