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>


int size;
#define OK 1
#define NG 0


/*size???????????\?????????????????§???????????????????????????????????£?????????????????§???????????§?????????????????????????????±?????????*/
void s_sort(int *S){
  int i,j,k,swap,count;
  if(size>=10){
    for(i=size-1;i>size-5;i--){
      j=0;
      while(j<=size-5){
        k=i;
        while(k>=j){
          if(S[k] < S[k-4]){
            swap=S[k];
            S[k]=S[k-4];
            S[k]=swap;
          }
          k-=4;
        }
        j+=4;    
      }
    }
  }
  for(i=0;i<size-1;i++){
    count=0;
    for(j=size-1;j>i;j--){
      if(S[j] <S[j-1]){
        swap=S[j];
        S[j]=S[j-1];
        S[j-1]=swap;
        count+=1;
      }
    }
    if(count==0) break;
  }
}


/*??????S???k?????????S[?????????]?????\????????°?????????????????????????????????????????????????????????????????¢??°*/
int judge(int *S,int k,int x){
  if(x==0) return OK;
  if(x < S[k]) return NG;
  else if(k==size){
    if(S[k]==x) return OK;
    else return NG;
  }
  if(judge(S,k+1,x-S[k]) +judge(S,k+1,x)) return OK;
  else return NG;
}




int main(){
  int i,*S,m,x;
  scanf("%d",&size);
  S=malloc(size*sizeof(int));
  for(i=0;i<size;i++) scanf("%d",&S[i]);
  s_sort(S);
  scanf("%d",&m);
  for(i=0;i<m;i++){
    scanf("%d",&x);
    if(judge(S,0,x)) printf("yes\n");
    else printf("no\n");
  }
  return 0;
}