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: Wrong Answer)

using System;

namespace B
{
    class Merge
    {
        public int count = 0;

        public void merge(ref int[] A, int left, int mid, int right)
        {
            int i, j=0, k;
            int n1 = mid - left;
            int n2 = right - mid;
            int[] L = new int[n1 + 1];
            int[] R = new int[n2 + 1];
            for (i = 0; i < n1; i++)
            {
                L[i] = A[left + i];
            }
            for (i = 0; i < n2; i++)
            {
                R[i] = A[mid + i];
            }
            L[n1] = int.MaxValue;
            R[n2] = int.MaxValue;
            i = 0;
            for (k = left; k < right; k++)
            {
                if (L[i] <= R[j])
                {
                    A[k] = L[i];
                    i++;
                    count++;
                }
                else
                {
                    A[k] = R[j];
                    j++;
                    count++;
                }
            }
            
        }

        public void mergeSort(ref int[] A, int left, int right)
        {
            if (left+1<right)
            {
                int mid = (left + right) / 2;
                mergeSort(ref A, left, mid);
                mergeSort(ref A, mid, right);
                merge(ref A, left, mid, right);
            }
        }
    }

    class B
    {
        static void Main(string[] args)
        {
            Merge sort = new Merge();
            string input;
            string[] input2 = new string[20000];
            int n, i;
            input = Console.ReadLine();
            n = int.Parse(input);
            int[] S = new int[n];
            input2 = Console.ReadLine().Split(' ');
            for (i = 0; i < n; i++)
            {
                S[i] = int.Parse(input2[i]);
            }

            sort.mergeSort(ref S, 0, n);
            for (i=0; i<n-1; i++)
            {
                Console.Write(S[i]+" ");
            }
            Console.WriteLine(S[n-1]);
            Console.WriteLine(sort.count);
        }
    }
}