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 Main {
	public static void main(String args[]) {
		Scanner scanner = new Scanner(System.in);
		int  numA = scanner.nextInt();//要素数
		int arrayA[] = new int[numA];
		
		for(int i = 0; i < numA; i ++ ){
			arrayA[i] = scanner.nextInt();
		}
		
		int  numM = scanner.nextInt();//要素数
		int arrayM[] = new int[numM];
		
		for(int i = 0; i < numM; i ++ ){
			arrayM[i] = scanner.nextInt();
		}
		
		Recursive(numA, arrayA, numM, arrayM);
		
		scanner.close();
	}
	
	public static void Recursive(int numA, int[] arrayA, int numM, int[] arrayM){ 
		
		for(int cou = 0; cou < numM; cou ++ ){
			if ( Recursive(numA, arrayA, 0, arrayM[cou]) ){
				System.out.println("yse");
			}else{
				System.out.println("no");
			}
		}
		
	}
	
	public static boolean Recursive(int numA, int[] arrayA, int i, int m){ 
		if( m == 0){
			return true;
		}
		
		if (i >= numA){
			return false;
		}
		
		return ( Recursive(numA,arrayA, i + 1, m)|| 
				Recursive(numA,arrayA,i + 1, m - arrayA[i]) );		//再帰
	
	}
}