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

int main()
{
   int na, nq, i, tf;
   int A[20], Q[200];

   i = 0;
   scanf("%d", &na);
   while (i < na) {
	  scanf("%d", &A[i]);
	  i++;
   }
   i = 0;
   scanf("%d", &nq);
   while (i < nq) {
	  scanf("%d", &Q[i]);
	  i++;
   }

   for (i = 0; i < nq; ++i) {
	  tf = solve(0, Q[i], na, A);
	  if (tf)
		 printf("yes\n");
	  else
		 printf("no\n");
   }
   return 0;
}