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: Time Limit Exceeded)

#include<stdio.h>

int main()
{
  int n;
  int a[20];
  int q;
  int m;
  int i, j, k;
  int kuri = 1;
  int kuritmp;
  int ans;
  int itmp;
  
  scanf("%d", &n);
  for(i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    kuri *= 2;
  }
  scanf("%d", &q);
  for(i = 0; i < q; i++) {
    scanf("%d", &m);
    for(k = 0; k < kuri; k++) {
      ans = 0;
      itmp = k;
      kuritmp = kuri;
      for(j = 0; j < n; j++) {
	kuritmp /= 2;
	if( kuritmp != 0 && itmp / kuritmp ) {
	  ans += a[j];
	  if(ans > m)
	    break;
	  itmp %= kuritmp;
	}
      }
      if(ans == m) {
	printf("yes\n");
	break;
      }
      if(k == kuri - 1) {
	printf("no\n");
      }
    }
  }
  return 0;
}