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;

namespace ALDS1_5_A
{
    class Program
    {
        static bool[,] vis = new bool[2001, 2001];
        static bool[,] ans = new bool[2001, 2001];
        static void Main(string[] args)
        {
            Console.ReadLine();
            var a = Array.ConvertAll(Console.ReadLine().Split(' '), int.Parse);
            Console.ReadLine();
            var m = Array.ConvertAll(Console.ReadLine().Split(' '), int.Parse);

            for (int i = 0; i < 2001; i++) vis[i, 0] = true;
            foreach (var q in m) Console.WriteLine(Solve(q, a.Length, a) ? "yes" : "no");
        }

        static bool Solve(int q, int n, int[] a)
        {
            if (q < 0) return false;
            if (q == 0) return true;
            if (vis[q, n]) return ans[q, n];
            for (int i = 0; i < n; i++)
            {
                var flag = Solve(q - a[i], i, a);
                if (flag)
                {
                    ans[q, n] = true;
                    break;
                }
            }
            vis[q, n] = true;
            return ans[q,n];
        }
    }
}