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.io.*;
import java.util.*;

class Main{
    public static void main(String[] args) throws Exception{
	int i=0;
	BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

	String str_n = br.readLine();
	String str_A = br.readLine();
	String str_q = br.readLine();
	String str_M = br.readLine();

	int n = Integer.parseInt(str_n);
	int q = Integer.parseInt(str_q);

	int[] A = new int[n];
	int[] M = new int[q];

	Scanner sc_A = new Scanner(str_A).useDelimiter("\\s* \\s*");
	Scanner sc_M = new Scanner(str_M).useDelimiter("\\s* \\s*");

	while(sc_A.hasNext()){
	    A[i] = sc_A.nextInt();
	    i++;
	}

	i = 0;//カウント初期化

	while(sc_M.hasNext()){
	    M[i] = sc_M.nextInt();
	    i++;
	}

	for(i=0;i<q;i++){
	    int m = M[i];
	    System.out.println(exhausitiveSearch(n,A,q,m));
	}

	sc_A.close();
	sc_M.close();
    }
    
    private static String exhausitiveSearch(int n,int A[], int q, int m){
	int flag = 0;
	
	for(int i=0;i<n;i++){
	    int COUNT = 0;
	    
	    if(m == A[i]){
		flag = 1;
		break;
	    }
	    
	    for(int j=i+1;j<n;j++){
		if((m == A[j]) || (m == A[i] + A[j])){
		    flag = 1;
		    break;
		}
		COUNT++;
	    }
	    if(COUNT == 1) break; 
	}
	
	if(flag == 1) return "yes";
	else return "no";
    }
}