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: Runtime Error)

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;

namespace ConsoleApplication1
{
    class Program
    {
        static void Main(string[] args)
        {
            int n = int.Parse(Console.ReadLine());
            int[] A = Array.ConvertAll(Console.ReadLine().Split(' '), int.Parse);
            int q = int.Parse(Console.ReadLine());
            int[] m = Array.ConvertAll(Console.ReadLine().Trim().Split(' '), int.Parse);

            StringBuilder sb = new StringBuilder();

            for (int i = 0; i < q; i++)
            {
                sb.AppendLine(SearchSum(A, n, m[i]) ? "yes" : "no");
            }
            Console.Write(sb);
        }

        static bool SearchSum(int[] A, int n, int m)
        {
            if (m == 0) return true;
            if (n == 0) return false;

            return SearchSum(A, n - 1, m) || SearchSum(A, n - 1, m - A[n - 1]);
        }
    }
}