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<string.h>
#include<stdlib.h>


int n,q;


//Aのｋ番目（0〜n-1）以降の要素をつかって、Mを作れるかを判定。成功：１　失敗:０
int judge(int *A,int M,int k){
  if(k==n-1){
    if(A[k]==M) return 1;
    else return 0;
  }
  else{
    if(M <A[k]) return judge(A,M,k+1);
    else if(M==A[k]) return 1;
    else{
      if(judge(A,M-A[k],k+1)==1 || judge(A,M,k+1)==1) return 1;
      else return 0;
    }
  }
}
  



int main(){
  int *A;
  int M;
  int i,j;
  
  scanf("%d",&n);
  A=malloc(n*sizeof(int));
  for(i=0;i<n;i++){
    scanf("%*[ \n]%d",&A[i]);
  }

  int max;
  int cha;
  for(i=0;i<n-1;i++){
    max=i;
    for(j=i+1;j<n;j++){
      if(A[max] <A[j]) max=j;
    }
    cha=A[max];
    A[max] =A[i];
    A[i]=cha;
  }//降順にソート

  
  scanf("%*[ \n]%d",&q);
  for(i=0;i<q;i++){
    scanf("%*[ \n]%d",&M);
    if(judge(A,M,0)==1) printf("yes\n");
    else printf("no\n");
  }
  return 0;
}