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: Wrong Answer)

import java.util.Scanner;

public class Main {
	int[] aArr;
	int aSize;
	int cnt;

	public static void main(String[] args) {
		new Main().run();
	}

	void run() {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		aArr = new int[n];
		for (int i = 0; i < n; i++) {
			aArr[i] = sc.nextInt();
		}
		aSize = n;
		int q = sc.nextInt();

		for (int i = 0; i < q; i++) {
			boolean ans = rec(0, sc.nextInt());
			if (ans) {
				System.out.println("yes");
			} else {
				System.out.println("no");
			}
		}
		sc.close();
	}

	boolean rec(int n, int m) {
		if (n >= aSize) {
			return false;
		}
		if (m < 0) {
			return false;
		}
		if (m == 0) {
			return true;
		}
		return (rec(n + 1, m) || rec(n + 1, m - aArr[n]));
	}
}