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: Runtime Error)

$stdin.gets.strip
A = $stdin.gets.strip.split(' ').map(&:to_i)
$stdin.gets.strip
ms = $stdin.gets.strip.split(' ').map(&:to_i)

def _solve(i, m)
  if m == 0
    return true
  end

  if i >= A.length
    return false
  end

  return solve(i+1, m) || solve(i+1, m - A[i])
end

@cache = {}
def solve(i, m)
  if res = @cache[[i, m]]
    res
  else
    @cache[[i, m]] = _solve(i, m)
  end
end

ms.each do |m|
  if solve(0, m)
    puts 'yes'
  else
    puts 'no'
  end
end