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: Accepted)

import java.util.Scanner;

class Main{

  static boolean solve(int[] a, int m, int i){
    if(m == 0)
      return true;
    if(i == a.length)
      return false;
    else
      return solve(a, m, i + 1) || solve(a, m - a[i], i + 1);
  }


  public static void main(String[] args){
    Scanner stdIn = new Scanner(System.in);
    int n = stdIn.nextInt();
    int[] a = new int[n];
    for(int i = 0; i < n; i++)
      a[i] = stdIn.nextInt();

    int q = stdIn.nextInt();
    int[] b = new int[q];
    for(int i = 0; i < q; i++)
      b[i] = stdIn.nextInt();

    for(int i = 0; i < q; i++){
      if(solve(a, b[i], 0))
        System.out.println("yes");
      else
        System.out.println("no");
    }
  }
}