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: Time Limit Exceeded)


import java.util.Scanner;

public class Main {

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

	private void run() {
		Scanner scanner = new Scanner(System.in);
		int n = scanner.nextInt();
		int[] num = new int[n];
		for (int i = 0; i < n; i++) {
			num[i] = scanner.nextInt();
		}
		int m = scanner.nextInt();
		while (m-- > 0) {
			boolean flag = false;
			int a = scanner.nextInt();
			for (int i = 0; i < 1 << n; i++) {
				int sum = 0;
				for (int j = 0; j < n; j++) {
					if (((i >> j) & 1) == 1) {
						sum += num[j];
					}
				}
				if (a == sum) {
					flag = true;
					break;
				}
			}
			System.out.println(flag ? "yes" : "no");
		}

	}
}