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 Ruby (judge verdict: Accepted)

# exhaustive.rb
#

s=Array.new(20)
table=Array.new(2000,0)
table[0]=1
table2=Array.new(2000,0)

n=STDIN.gets.split[0].to_i
str=STDIN.gets.split
sum=0
0.upto(n-1) {|i|
  table2=Marshal.load(Marshal.dump(table))
  num = str[i].to_i
  0.upto(sum){|j|
    break if j+num > 2000 
    if table[j]==1
      table2[j+num]=1 
#     printf("%d %d %d\n",j,num,j+num)
    end
  } 
  sum += num
  table=Marshal.load(Marshal.dump(table2))
}

q  =STDIN.gets.split[0].to_i
str=STDIN.gets.split

0.upto(q-1){|i|
  m=str[i].to_i
  puts (table[m]==1)?"yes":"no"
}