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: Runtime Error)

import itertools
n=int(input())
A=[0 for i in range(n)]
A=input().split()
for i in range(n):
    A[i]=int(A[i])
q=int(input())
B=[0 for i in range(q)]
B=input().split()
C=[]
for i in range(q):
    B[i]=int(B[i])
l=[0,1]
p=[]
k=0
D=[]
for v in itertools.product(l,repeat=n):
    sum=0
    for i in range(n):
        if v[i]==1:
            sum=sum+A[i]
    D.append(sum)
    k+=1
for i in range(q):
    c=0
    for j in range(k):
        if B[i]==D[j] and c==0:
            C.append("yes")
            c+=1
    if c==0:
        C.append("no")
for i in range(q):
    print(C[i])