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;
using System.Linq;
using System.Text;
using System.Threading.Tasks;

namespace Aoj_Learning2
{
    class Program
    {
        static void Main(string[] args)
        {
            Aoj_Learning17();
        }

        static void Solve(ref int[] S, int i, int size, ref bool[] ans, int[] Sequence1, int[] Sequence2, int Size2)
        {
            if (size == i)
            {
                int sum = 0;
                for(int j = i-1; j >= 0; j--)
                {
                    if(1 == S[j])
                    {
                        sum += Sequence1[j];
                    }
                }
                // 数列内まとめて確認
                for (int j = 0; j < Size2; j++)
                {
                    if (Sequence2[j] == sum) ans[j] = true;
                }
                return;
            }

            Solve(ref S, i + 1, size, ref ans, Sequence1, Sequence2, Size2);
            S[i] = 1;
            Solve(ref S, i + 1, size, ref ans, Sequence1, Sequence2, Size2);
            S[i] = 0;
        }

        private static void Aoj_Learning17()
        {
            string[] str;
            int iSequenceCount1 = 0, iSequenceCount2 = 0;
            int[] iSequence1 = new int[20];
            int[] iSequence2 = new int[200];
            int[] iSelect = new int[20];

            // 数列を取得
            iSequenceCount1 = int.Parse(Console.ReadLine());
            str = Console.ReadLine().Split(' ');
            for(int i = 0; i < iSequenceCount1; i++)
            {
                iSequence1[i] = int.Parse(str[i]);
            }
            iSequenceCount2 = int.Parse(Console.ReadLine());
            str = Console.ReadLine().Split(' ');
            for (int i = 0; i < iSequenceCount2; i++)
            {
                iSequence2[i] = int.Parse(str[i]);
            }

            // 数列2内の数値を数列1内の数値の和で表現可能か探索
            bool[] b = new bool[iSequenceCount2];
            // 再起処理
            Solve(ref iSelect, 0, iSequenceCount1, ref b, iSequence1, iSequence2, iSequenceCount2);

            for (int i = 0; i < iSequenceCount2; i++)
            {
                if (true == b[i])
                {
                    Console.WriteLine("yes");
                }
                else
                {
                    Console.WriteLine("no");
                }
            }           
            
        }
    }
}