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 Go (judge verdict: Accepted)

package main

import "fmt"

var S, T [100000]int
var n, q int

func main() {
	var m int
	fmt.Scan(&n)
	for i := 0; i < n; i++ {
		fmt.Scan(&S[i])
	}
	fmt.Scan(&q)
	for i := 0; i < q; i++ {
		fmt.Scan(&m)
		if solve(0, m) == true {
			fmt.Println("yes")
		} else {
			fmt.Println("no")
		}
	}
}
func solve(i, m int) bool {
	if m == 0 {
		return true
	}
	if i >= n {
		return false
	}
	if solve(i+1, m) == true {
		res := solve(i+1, m)
		return res
	} else if solve(i+1, m-S[i]) == true {
		res := solve(i+1, m-S[i])
		return res
	}
	return false
}