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 Haskell (judge verdict: Time Limit Exceeded)

import Data.List
import Data.List.Split

strToInt str = (read str :: Int)
getLineInt = fmap strToInt getLine
getLineInts = fmap ((map strToInt) . (splitOn " ")) getLine

solve :: [Int] -> Int -> Int -> Int -> Bool
solve ls n summed k =
  if ((ls !! n) + summed == k) then
    True
  else if (((length ls) - 1) > n) then
    (solve ls (n+1) summed k) || (solve ls (n+1) (summed + (ls !! n)) k)
  else
    False

main = do
  n <- getLineInt
  xs <- getLineInts
  m <- getLineInt
  sums <- getLineInts
  let set = (solve xs)
  putStrLn $ intercalate "\n" $ map (\x -> if x then "yes" else "no") (map (set 0 0) sums)