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 Python3 (judge verdict: Time Limit Exceeded)

# -*- coding:utf-8 -*-
import sys


def exhaustive(n, target, lst):
    if n == target:
        return True
    if n > target or lst == []:
        return False
    else:
        for i, val in enumerate(lst, 1):
            if exhaustive(n + val, target, lst[i:]):
                return True
    return False


def main(data, targets):
    for target in targets:
        if exhaustive(0, target, data):
            print("yes")
        else:
            print("no")


if __name__ == "__main__":
    lst = [val.split() for val in sys.stdin.read().splitlines()]
    data = [int(n) for n in lst[1]]
    targets = [int(n) for n in lst[3]]
    main(data, targets)