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 Ruby (judge verdict: Accepted)

require 'set'
num_of_n = gets.chomp.to_i
n_list = gets.chomp.split.map(&:to_i)

num_of_m = gets.chomp.to_i
m_list = gets.chomp.split.map(&:to_i)


def main(num_of_n, n_list, num_of_m, m_list)
  sum_list = [].to_set

  (1..2**num_of_n).each do |n_ten|
    n_two = n_ten.to_s(2).rjust(num_of_n, '0')
    sum = n_two.chars.map(&:to_i).map.with_index{ |flag, index| flag == 1 ? n_list[index] : 0 }.sum

    sum_list.add(sum)
  end

  m_list.each do |m|
    if sum_list.include?(m)
      puts 'yes'
    else
      puts 'no'
    end
  end
end

main(num_of_n, n_list, num_of_m, m_list)