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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

public class Main {
	public static void main(String[] args) throws IOException {
		new Main().run();
	}
	private void run() throws IOException {
		BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
		int n = Integer.parseInt(reader.readLine());
		int[] a = new int[n];
		StringTokenizer st = new StringTokenizer(reader.readLine());
		for (int i = 0; i < n; i++) {
			a[i] = Integer.parseInt(st.nextToken());
		}
		int q = Integer.parseInt(reader.readLine());
		st = new StringTokenizer(reader.readLine());
		for (int i = 0; i < q; i++) {
			if (subsetSum(Integer.parseInt(st.nextToken()), a)) {
				System.out.println("yes");
			} else {
				System.out.println("no");
			}
		}
		reader.close();
	}

	private boolean subsetSum(int s, int[] a) {
		int n = a.length;
		boolean[][] dp = new boolean[n+1][s+1];
		for (int i = 1; i <= n; i++) {
			int v = a[i-1];
			for (int j = 0; j <= s; j++) {
				if (v > j) {
					dp[i][j] = dp[i-1][j];
				} else if (v == j) {
					dp[i][j] = true;
				} else {
					dp[i][j] = dp[i-1][j] || dp[i-1][j-v];
				}
			}
		}
		return dp[n][s];
	}
}