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>
#include <stdlib.h>


int solve(int i, int m, int *a, int n) {
  //solve( 0, 探したい数, 配列のポインタ, 配列の長さ);


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


int *inputIntArray(int n) { 

  int *a;
  int i;

  a = (int *)malloc(sizeof(int) *n);

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

  return a; //配列ポインタ

}


int main() {

  int n,q;
  int *a, *m;

  int i;

  scanf("%d", &n);
  a = inputIntArray( n );

  scanf("%d", &q);
  m = inputIntArray( q );

  for( i=0; i<q; i++) {

    if( solve( 0, m[i], a, n) ) {
      printf("yes\n");
    }
    else {
      printf("no\n");
    }

  }

    return 0;

}