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)

def can_make(a, q)
  if q == 0
    return true
  elsif a.nil? == true || a.empty? == true
    return false
  elsif a.length == 1
    if q == a[0]
      return true
    else
      return false
    end
  else
     can_make(a[1..-1], q-a[0]) || can_make(a[1..-1], q)
  end
end

n = gets.chomp.to_i
a = gets.chomp.split.map(&:to_i)
q = gets.chomp.to_i
m = gets.chomp.split.map(&:to_i)


for i in 0..(q-1)
  if can_make(a, m[i]) == true
    puts "yes"
  else
    puts "no"
  end
end