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 = int(input())
A = [int(i) for i in input().split()]
q = int(input())
m = [int(i) for i in input().split()]
    
def zentansaku(j,k,l):  # ??????A[???]???????????? A[j]???A[l?????????A[N??????] ?????§????¶????????????????
     num = 0
     for i in range(l,N-1):
         Z = j + A[i]
         
         if Z == k:
             return True
         else :
             if j+1 == N :
                 return False
             else :
                 zentansaku(Z,k,l+1)
         num += 1        
                

for i in range(q):
    check = 0
    for i2 in range(N-1):
        if zentansaku(A[i2],m[i],i2+1) == True:
            print("yes")
            check = 1
    if check == 0 :
        print("no")