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.Collections.Generic;
using System.Linq;
class ilds1_5_a{
  public static void Main(){
    int n = int.Parse(Console.ReadLine());
    int[] a = Console.ReadLine().Split(' ').Select(s => int.Parse(s)).ToArray();
    int q = int.Parse(Console.ReadLine());
    int[] m = Console.ReadLine().Split(' ').Select(s => int.Parse(s)).ToArray();
    for(int i=0; i<q; i++){
      if(sumup(0,a,q,m[i],n)){
        Console.WriteLine("yes");
      }
      else{
        Console.WriteLine("no");
      }
    }
  }
  public static bool sumup(int t, int[] a, int q, int cur, int size){
    if(cur == 0){
      return true;
    }
    if(t == size){
      return false;
    }
    return sumup(t+1, a, q, cur, size) || sumup(t+1, a, q, cur-a[t], size);
  }
}