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)

count = 0
input_a = ""
input_m = ""
STDIN.each_line {|line|
  if count == 0 || count == 2
    count += 1
    next
  end

  if count == 1
    input_a = line
  end

  if count == 3
    input_m = line
  end
  count += 1
}


$input_a_list = input_a.split(" ")
$n = $input_a_list.length
input_m_list = input_m.split(" ")
$check = {}
$memo = {}
def solve i, m
  if i == $n
    return m == 0
  end

  if !$memo.has_key? (i + 1)
    $memo[i + 1] = {}
  end

  res1 = false
  if $memo[i + 1].has_key? (m)
    res1 = $memo[i + 1][m]
  else
    res1 = solve i + 1, m
    $memo[i + 1][m] = res1
  end

  input_a = $input_a_list[i].to_i
  res2 = false
  if (m - input_a) < 0
    return (res1 | res2)
  end

  if $memo[i + 1].has_key? (m - input_a)
    res2 = $memo[i + 1][m - input_a]
  else
    res2 = solve i + 1, m - input_a
    $memo[i + 1][m - input_a] = res2
  end

  return (res1 | res2)
end

input_m_list.each  {|num|
  if solve 0, num.to_i
    puts "yes"
  else
    puts "no"
  end
}