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;

public static class Program
{
    public static void Main()
    {
        new Solve().Run();
    }
}

public class Solve
{
    int[] a;
    int[] query;
    int n;

    public void Run()
    {
        Console.ReadLine();
        a = Console.ReadLine().Split(' ').Select(x => int.Parse(x)).ToArray();
        n = a.Length;
        Console.ReadLine();
        query = Console.ReadLine().Split(' ').Select(x => int.Parse(x)).ToArray();
        foreach (var m in query)
        {
            Console.WriteLine(Func2(0, 0, m) ? "yes" : "no");
        }
    }

    public bool Func2(int i, int sum, int m)
    {
        if (i == n)
        {
            return sum == m;
        }
        return Func2(i + 1, sum, m) || Func2(i + 1, sum + a[i], m);
    }
}