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: Wrong Answer)

gets
a = gets.split(' ').map(&:to_i).sort.reverse
gets
b = gets.split(' ').map(&:to_i)

max = b.max
count = 0

loop do
  pat = format("%0#{a.size}d", count.to_s(2))
  break if pat.size > a.size
  sum = 0
  pat.size.times { |x| sum += (pat[x] == '0' ? 0 : a[x])  }
  break if sum > max
  while b.include?(sum)
    b[b.index(sum)] = 'yes'
  end
  count += 1
end

b.each { |e| puts e == 'yes' ? e : 'no' }