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>
#include<stdlib.h>
#include<math.h>
int kaijou(int n, int m);
int S[50],n1;
int main(){
  int n2,m;
  int i;
 
  scanf("%d",&n1);
  
 
 for(i=0; i<n1; i++){
   scanf("%d",&S[i]);
 }

 
 scanf("%d",&n2);

 for(i=0; i<n2; i++){
   scanf("%d",&m);
   if(kaijou(0,m)){
     printf("yes\n");
   }

   else{
     printf("no\n");
 }
 }

   return 0;
 }

   


int kaijou(int i,int m){

  if(m==0){
    return 1;
  }
  if(i>=n1){
    return 0;
  }


  return kaijou(i+1,m)||kaijou(i+1,m-S[i]);
  
 }