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.Generic;
using System.Linq;

namespace Recursion___Divide_and_Conquer
{
    class Program
    {
        static Queue<int> queue;
        static int sum;

        static void Main(string[] args)
        {
            queue = new Queue<int>();
            Console.ReadLine();
            var A = Console.ReadLine().Split(' ').Select(s => int.Parse(s)).ToArray();

            for (int i = 0; i < A.Length; i++)
            {
                queue.Enqueue(A[i]); sum += A[i];
            }

            Console.ReadLine();
            var m = Console.ReadLine().Split(' ').Select(s => int.Parse(s)).ToArray();
            for (int i = 0; i < m.Length; i++)
                if (CanMake(m[i])) Console.WriteLine("yes");
                else Console.WriteLine("no");
        }

        static bool CanMake(int n)
        {
            if (n < 0 || n > sum) return false;
            if (n == 0 || queue.Contains(n)) return true;
            var c = queue.Count;
            for (int i = 0; i < c; i++)
            {
                var m = queue.Dequeue();
                if (CanMake(n - m)) { queue.Enqueue(m); return true; }
                queue.Enqueue(m);
            }
            return false;
        }
    }
}