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 JAVA (judge verdict: Runtime Error)

import java.util.Arrays;
import java.util.HashSet;
import java.util.Scanner;
import java.util.Set;

//??????a[]?????¨??????????¶??????????????????°??????
  
public class Main {

  static int n ;
  
  static int a[] = new int[405];

  static  Set<Integer> d=new HashSet<Integer>();
  static int sum[] = new int[32];
  
  static int sum_idx; 

  static int m;

  //????????¢??°?????§a[cnt]????¶????????????????tmp = ????????£?????????
  public static void dfs(int cnt, int tmp) {


    if (cnt == n) {
      sum[sum_idx] = tmp;
      d.add(sum[sum_idx]);
      sum_idx++;
     
      m++;
      return; 
    }

//    System.out.println("(not exit)cnt, tmp = " + String.valueOf(cnt) + ", " + String.valueOf(tmp));

    dfs(cnt+1, tmp);
    dfs(cnt+1, tmp + a[cnt]);
  }

  public static void main(String[] args) {
	Scanner sc=new Scanner(System.in);
	n=Integer.parseInt(sc.next());
	for(int i=0;i<n;i++){
		a[i]=Integer.parseInt(sc.next());
	}
	
    m = 0;
    sum_idx = 0;
    Arrays.fill(sum, 0);
    dfs(0, 0);
    
   
   
    
    int q=Integer.parseInt(sc.next());
    
    int x;
    for(int j=0;j<q;j++)
    {
    	x=Integer.parseInt(sc.next());
    
    	System.out.println((d.contains(x)) ? ("yes") : ("no"));
    }
    
  }
}