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)

def solve2(arr)
  ans = { }
  ans[[0, 0]] = true
  limit = arr.inject(0) { |s, e| s + e}
  arr.each_with_index do |elem, ind|
    (0..limit).each do |sum|
      if ans[[ind, sum]] or ans[[ind, sum - elem]]
        ans[[ind + 1, sum]] = true
      end
    end
  end
  p ans
  ans  
end

def main2
  n = gets.chomp.to_i
  arr = gets.chomp.split(/\s/).collect { |x| x.to_i}
  q = gets.chomp.to_i
  questions = gets.chomp.split(/\s/).collect { |x| x.to_i}
  ans = solve2(arr)
  questions.each do |question|
    if ans[[arr.size, question]]
      puts "yes"
    else
      puts "no"
    end
  end  
end

main2