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 {
  public static void main(String[] args) {
    Scanner in = new Scanner(System.in);    
    int n = in.nextInt();
    int[] A = new int[n];
    for (int i = 0; n != 0; i++) {
      A[i] = in.nextInt();
      n--;
    }
     
    int q = in.nextInt();
    int[] M = new int[q];
    for (int i = 0; q != 0; i++) {
      M[i] = in.nextInt();
      q--;
    }
     
    for (int i = 0; i < M.length; i++) {
      System.out.println((dfs(0, A, 0, M[i])) ? "yes" : "no" );
    }
  }
  static boolean dfs(int p, int[] A, int t, int sum) {
    if (t < 0 ) return false;
    if (p == A.length) return t == sum;
    if (dfs(p + 1, A, t + A[p], sum)) return true;
    if (dfs(p + 1, A, t, sum))        return true;
    return false;
  }
}