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: Wrong Answer)

import scala.io.StdIn.{readInt, readLine};
object Main {
    def main(args: Array[String]): Unit = {
        readInt;
        val arr = readLine.split(" ").map(_.toInt);
        readInt;
        val es = new ExhaustiveSearch(arr);
        readLine.split(" ").map(_.toInt).foreach { target => if (es.solve(target)) println("yes") else println("no")}
    }
}
class ExhaustiveSearch(arr: Array[Int]) {
    val size = arr.length;
    val sum = arr.foldLeft(0){ (s, v) => s + v};
    val dpArr = Array.ofDim[Int](size, sum + 1);
    def solve(target: Int): Boolean = {
        def _solve(idx: Int, t: Int): Boolean = {
            if (t < 0 || t > sum || idx >= size) {
                false;
            } else {
                if (dpArr(idx)(t) == 1) {
                    true;
                } else if (dpArr(idx)(t) == -1) {
                    false;
                } else {
                    if (t == 0 || _solve(idx + 1, t) || _solve(idx + 1, t - arr(idx))) {
                        dpArr(idx)(t) = 1;
                        true;
                    } else {
                        dpArr(idx)(t) = -1;
                        false;
                    }
                }
            }
        }
        _solve(0, target);
    }
}