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
int solve(int , int );

int n,A[100*N];

int main()
{
  int i,q,m[10*N],sum,hantei;
  
  while(1)
    {
      scanf("%d",&n);
      if(n<=N) break;
    }

  i=0;
  while(i<n)
    {
      scanf("%d",&A[i]);
      if(A[i] >= 1 && A[i] <= 100*N) i++;
    }

  while(1)
    {
      scanf("%d",&q);
      if(q<=10*N) break;
    }

  i=0;
  while(i<q)
    {
      scanf("%d",&m[i]);
      if(m[i] >= 1 && m[i] <= 100*N) i++;
    }
  for(i=0;i<q;i++)
    {
      sum = m[i];
      hantei = solve(0,sum);
      if(hantei == 1) printf("yes\n");
      else printf("no\n");
    }
      
  return 0;
}


int solve(int i, int m)
{
  int res;
  if(m == 0) return 1;
  if (i >= n) return 0;
  if(solve(i+1,m) || solve(i+1,m-A[i])) res = 1;
  else res = 0;
  return res;
}