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: Accepted)

import sys
input = sys.stdin.readline
from operator import itemgetter
sys.setrecursionlimit(10000000)
INF = 10**30

def main():
    n = int(input().strip())
    a = list(map(int, input().strip().split()))
    q = int(input().strip())
    m = list(map(int, input().strip().split()))
    l = []
    for i in range(2**n):
        w = 0
        for j in range(n):
            if 1 & (i >> j) == 1:
                w += a[j]
        l.append(w)
    l = set(l)
    for i in range(q):
        if m[i] in l:
            print('yes')
        else:
            print('no')
if __name__ == '__main__':
    main()