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 N 20
#define Q 2000

//int rec(int,int [N],int [N],int,int,int);
int solve(int ,int ,int ,int [N],int);

int main(){
  int num,n[N],qu,q[Q],i,S[N];

  scanf("%d",&num);

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

  scanf("%d",&qu);

  for(i=0;i<qu;i++) scanf("%d",&q[i]);

  for(i=0;i<qu;i++){
    //if(rec(0,S,n,q[i],num,0)==1) printf("yes\n");
    if(solve(0,q[i],num,n,0)==1) printf("yes\n");
    else printf("no\n");
  }
  return 0;
}



/*
int rec(int i,int S[N],int n[N],int q,int num,int flag){
  int j,sum=0;
  if(i==num){
    if(flag==0){
      for(j=0;j<num;j++){
	if(S[j]==1)sum+=n[j];
	if(sum>=q)break;
      }
      if(sum==q) flag=1;
    }
    return flag;
  }
  
  S[i]=1;
  flag=rec(i+1,S,n,q,num,flag);
  S[i]=0;
  flag=rec(i+1,S,n,q,num,flag);

  return flag;
}
*/

int solve(int i,int q,int num,int n[N],int res){
  if(q==0 || res==1) return 1;
  if(i>=num) return 0;
  res = solve(i+1,q,num,n,res);
  res = solve(i+1,q-n[i],num,n,res);
  return res;
}