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 Scala (judge verdict: Accepted)

import scala.io.StdIn._
import scala.collection.mutable

object Main extends App {
  
  def solve(m: Int, a: Array[Int], i: Int = 0, memo: mutable.Map[(Int, Int), Boolean]= mutable.Map()): Boolean = m match {
    case 0 => true
    case _ if i >= a.size => false
    case _ if (memo.get(i, m) != None) => memo((i, m))
    case _ => 
      val r = solve(m, a, i + 1, memo) || solve(m - a(i), a, i + 1, memo)
      memo += ((i, m) -> r)
      r
  }

  val N = readLine.toInt
  val A = readLine.split(" ").map(_.toInt)
  val Q = readLine.toInt
  val M = readLine.split(" ").map(_.toInt)
  M.foreach(m => if (solve(m, A)) println("yes") else println("no"))
}