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.Arrays;
import java.util.Scanner;

public class Main {
	private static int n, m;
	private static int[] a, b;
	private static int[][] dp;
	
	public static void main(String[] args) {
		inputData();
		solve();
	}
	
	private static void inputData() {
		try (Scanner scan = new Scanner(System.in)) {
			n = scan.nextInt();
			a = new int[n];
			for(int i=0; i<n; i++) a[i] = scan.nextInt();
			m = scan.nextInt();
			b = new int[m];
			for(int i=0; i<m; i++) b[i] = scan.nextInt();
		}
		dp = new int[n+1][2000+1];
		for(int[] row : dp) Arrays.fill(row, -1);
	}
	
	private static void solve() {
		for(int i=0; i<m; i++) {
			if(serchNumber(0, b[i]) == 1) {
				System.out.println("yes");
			} else {
				System.out.println("no");
			}
		}
	}
	
	private static int serchNumber(int i, int m) {
		if(m < 0) return 0;
		if(dp[i][m] != -1) return dp[i][m];
		     if(m == 0)                        dp[i][m] = 1;
		else if(i == n)                        dp[i][m] = 0;
		else if(serchNumber(i+1, m) == 1)      dp[i][m] = 1;
		else if(serchNumber(i+1, m-a[i]) == 1) dp[i][m] = 1;
		else                                   dp[i][m] = 0;     
		return dp[i][m];     
	}
}