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;


public class Main {

	public static boolean isSolved;
	public static int[] A;
	public static int n;
	
	static void solve(int index, int num) {
		if (num == 0) {
			isSolved = true;
			return;
		}

		if (isSolved || num < 0 || index == n) return;
		
		solve(index+1, num - A[index]);
		solve(index+1, num);
	}
	
	public static void main(String[] args) {
		Scanner scan = new Scanner(System.in);

		n = scan.nextInt();
		A = new int[n];
		for (int i = 0; i < n; i++) {
			A[i] = scan.nextInt();
		}
		
		int q = scan.nextInt();
		int[] M = new int[q];
		for (int i = 0; i < q; i++) {
			M[i] = scan.nextInt();
		}
		
		for (int i = 0; i < q; i++) {
			isSolved = false;
			
			solve(0, M[i]);
			
			if (isSolved) System.out.println("yes");
			else System.out.println("no");
		}
		
		scan.close();
	}

}