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 Python (judge verdict: Wrong Answer)

'''
Created on 2016/03/18

@author: toshi
'''

#n = map(int, raw_input())
n = int(raw_input())
A = map(int, raw_input().split())
#q = map(int, raw_input())
q = int(raw_input())
m = map(int, raw_input().split())

A = sorted(A)

def sum_A(A, n, pos, sum, m, res):
    if res[0] == "yes":
        return 0
    if sum == m:
        res[0] = "yes"
        return 0
    if pos + 1 > n - 1:
        return 0
    sum_A(A, n, pos + 1, sum + A[pos], m, res)
    sum_A(A, n, pos + 1, sum, m, res)

for i, m_i in enumerate(m):
    sum = 0
    res = ["no"]
    sum_A(A, n, 0, sum, m_i, res)
    print res[0]