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: Wrong Answer)

n, a, q, ms = int(input()), list(map(int, input().split())), input(), map(int, input().split())

cache = {}


def sum_a(tmp, bin, i=0, mask=1):
    global a
    if not bin:
        return tmp
    elif bin in cache:
        return cache[bin]
    elif bin & mask:
        tmp += a[i]
    result = sum_a(tmp, bin & ~mask, i + 1, mask << 1)
    cache[bin] = result
    return result


for m in ms:
    for bin in range(1, 2 ** n - 1):
        if sum_a(0, bin) == m:
            print('yes')
            break
    else:
        print('no')