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;

class Solve{
	public int plus[];
	public int ans[];
	public int count;
	public void printans(int a,int b,int i) { 
		if(b == 0) System.out.printf("yes\n");
		else {
			count++;
			if(a < plus.length) {
				printans(a + 1,b - plus[a],i);
				printans(a + 1,b,i);
			}
		}
	}
}

class Main{
	public static void main(String[] args) 	{
		Scanner sc = new Scanner(System.in);
		Solve sv = new Solve();
		//prepare facts
		int num = sc.nextInt();
		sv.plus = new int[num];
		for(int i = 0;i < num;i++) {
			sv.plus[i] = sc.nextInt();
		}
		//prepare answers
		int ansnum = sc.nextInt();
		sv.ans = new int[ansnum];
		for(int i = 0;i < ansnum;i++) {
			sv.ans[i] = sc.nextInt();
		}
		//print yes or no
		for(int i = 0;i < ansnum;i++) {
			sv.count = 0;
			sv.printans(0,sv.ans[i],i);
			if(sv.count == Math.pow(2,sv.plus.length + 1) - 1) System.out.printf("no\n");
		}
		sc.close();
	}
}