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)

n = input()
A = input().split(' ')
A.sort()
q = input()
M = input().split(' ')

A = list(map(int,A))
M = list(map(int,M))

#memo = [True]*(len(A))

def search(base, m, n):
    if base == m:
        return True
    result = False
    for i in range(n,len(A)):
        if base + A[i] <= m: # and memo[i]:
#           memo[i] = False
#           print(str(A[i]) + ',' + str(base+A[i]) + ',')
            result = result or search(base+A[i], m, i+1)
#           memo[i] = True
    return result

def print_yn(result):
    if result:
        print('yes')
    else:
        print('no')

for m in M:
#   print(str(m)+':')
    print_yn(search(0,m,0))