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;

class A5{
  static int n;
  static string[] str;
  static int[] A;
  static int q;
  static string[] str2;
  static int[] M;

  public static void Main(){
    n = int.Parse(Console.ReadLine());
    str = Console.ReadLine().Split(' ');
    A = new int[n];
    q = int.Parse(Console.ReadLine());
    str2 = Console.ReadLine().Split(' ');
    M = new int[q];


    for(int i = 0;i < n;i++){
      A[i] = int.Parse(str[i]);
    }
    for(int i = 0;i < q;i++){
      M[i] = int.Parse(str2[i]);
    }

    for(int i = 0;i < q;i++){
      if(Solve(0,M[i]))
       Console.WriteLine("yes");
      else Console.WriteLine("no");
    }
}

  static bool Solve(int i,int m){
      if(m == 0) return true;
      if(i >= n) return false;
      bool res = Solve(i + 1,m)|| Solve(i + 1,m - A[i]);
      return res;
    }
}