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.collection.mutable
import scala.io.StdIn.{readInt, readLine}

object Main extends App {

  val n  = readInt()
  val xs = readLine().split(" ").map { _.toInt }
  readLine()
  val ys = readLine().split(" ").map { _.toInt }

  val zs = mutable.Set.empty[Int]
  for {
    i  <- 1 to n
    cs <- xs.combinations(i).map { _.sum }
  } {
    zs += cs
  }

  for {
    y  <- ys
    ans = if (zs.contains(y)) "yes" else "no"
  } {
    println(ans)
  }
}