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_5A_2{
    public class Program{
        public static void Main(string[] args){
            var sr = new StreamReader();
            //---------------------------------
            var N = sr.Next<int>();
            var A = sr.Next<int>(N);
            var Q = sr.Next<int>();
            var M = sr.Next<int>(Q);

            Func<int, int, bool> solve = null;
            solve = (i, x) =>{
                if(x == 0) return true;
                if(i >= N) return false;
                return solve(i + 1, x - A[i]) || solve(i + 1, x);
            };

            foreach(var m in M){
                Console.WriteLine(solve(0, m) ? "yes" : "no");
            }
            //---------------------------------
        }
    }

    public class StreamReader{
        private readonly char[] _c = {' '};
        private int _index = -1;
        private string[] _input = new string[0];

        public T Next<T>(){
            if(_index == _input.Length - 1){
                _index = -1;
                while(true){
                    string rl = Console.ReadLine();
                    if(rl == null){
                        if(typeof(T).IsClass) return default(T);
                        return (T)typeof(T).GetField("MinValue").GetValue(null);
                    }
                    if(rl != ""){
                        _input = rl.Split(_c, StringSplitOptions.RemoveEmptyEntries);
                        break;
                    }
                }
            }
            return (T)Convert.ChangeType(_input[++_index], typeof(T), System.Globalization.CultureInfo.InvariantCulture);
        }

        public T[] Next<T>(int x){
            var ret = new T[x];
            for(var i = 0; i < x; ++i) ret[i] = Next<T>();
            return ret;
        }

        public T[][] Next<T>(int y, int x){
            var ret = new T[y][];
            for(var i = 0; i < y; ++i) ret[i] = Next<T>(x);
            return ret;
        }
    }
}