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 Kotlin (judge verdict: Accepted)


var n:Int = 0
var A: MutableList<Int> = mutableListOf<Int>()
fun main(args: Array<String>) {
    n = readLine()!!.toInt()
    A.addAll(readLine()!!.split(" ").map { it.toInt() })
    val q = readLine()!!.toInt()
    val M: List<Int> = readLine()!!.split(" ").map { it.toInt() }
    //MをAから作成可能か判定
    for (i in (0..q-1))
    {
        if (solve(0,M[i])==1)
        {
            println("yes")
        }
        else
        {
            println("no")
        }
    }

}
private fun solve(i:Int,m:Int):Int
{
    if (m==0)
    {
        return 1
    }
    if (m < 0)
    {
        return 0
    }
    if(i >= n)
    {
        return 0
    }

    if ((solve(i+1,m)==1) || (solve(i+1,m-A[i])==1))
    {
        return 1
    }
    else
    {
        return 0
    }
}