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 a 2000
#define b 2000

int A[a], Mi[b];
int hantei(int, int);

int main()
{
  int n, q, i, j, k;
  // int A[a], Mi[b];

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

  scanf("%d", &q);
  for(j=0; j<q; j++)
    {
      scanf("%d", &Mi[j]);
    }

  for(k=0; k<q; k++)
    {
      if(hantei(n, Mi[k]))
	{
	  printf("yes\n");
	}
      else
	{
	  printf("no\n");
	}	
    }

  return 0;
}

int hantei(int v, int w)
{
  int x;
  if(w == A[v])
    {
      return 1;
    }
  if(v == -1)
    {
      return 0;
    }
  if(w <= 0)
    {
      return 0;
    }
  if(hantei(v-1,w))
    {    
      return 1;
    }
  if(hantei(v-1,w-A[v]))
    {
      return 1;
    }

  return 0;   
}