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)

n = int(input())
aaa = list(map(int, input().split()))
q = int(input())
mmm = list(map(int, input().split()))
suma = sum(aaa)
def solve(i,m):
  global n,aaa,suma
  if m<0 or m>suma:
    return False
  if m==0:
    return True
  if i >= n:
    return False
  return solve(i+1,m) or solve(i+1,m-aaa[i])
mina = min(aaa)
for m in mmm:
  if m < mina or m>suma:
    print("no")
  else:
    if solve(0,m):
      print("yes")
    else:
      print("no")