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 C (judge verdict: Accepted)

#include <stdio.h>
#define true 1
#define false -1
#define M 2000

int x[M],y[200],n;

int solve (int i,int j)
{
  int res;
  if(j==0) return true;
  if(i>=n) return false;
  if(solve(i+1,j)==true) res=true;
  else if(solve(i+1,j-x[i])==true)res=true;
  else res=false;
  return res;
}

int main()
{
  int a,b,c,d=0;
  char s[2][4]={"yes","no"};

  scanf("%d",&n);
  for(a=0;a<n;a++)scanf("%d",&x[a]);

   scanf("%d",&b);
   for(a=0;a<b;a++){
      scanf("%d",&y[a]);
      if(solve(0,y[a])==true)printf("%s\n",s[0]);
      else printf("%s\n",s[1]);
   }
   return 0;
}