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.

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].

<image>

Constraints

* n ≤ 20
* q ≤ 200
* 1 ≤ elements in A ≤ 2000
* 1 ≤ Mi ≤ 2000

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.

Example

Input

5
1 5 7 10 21
8
2 4 17 8 22 21 100 35


Output

no
no
yes
yes
yes
yes
no
no

import java.util.Scanner;

class Main {

	public static void main(String args[]) {
		Scanner in = new Scanner(System.in);
		int n = in.nextInt();

		int[] c = new int[n];
		for (int i = 0; i < n; i++) {
			c[i] = in.nextInt();

		}
		int q = in.nextInt();
		for (int i = 0; i < q; i++) {

			if(solve(c,0,in.nextInt())) {
				System.out.println("yes");
			}else{
				System.out.println("no");
			};

		}

		in.close();
	}

	private static boolean solve(int[] c,int i, int m) {
		boolean res;
	   if (m == 0) {
	     return true;
	   }
	   if (i == c.length) {
	     return false;
	   }
	   res = solve(c,i + 1, m) || solve(c,i + 1, m - c[i]);
	   return res;
	}
}

