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.*;


//I have to use hashset
public class Main{
    public static void main(String[] args)throws IOException{
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        String str;
        
        int n = Integer.parseInt(br.readLine());
        str = br.readLine();

        String[] A = str.split("\\s");
        int q = Integer.parseInt(br.readLine());
        str = br.readLine();

        String[] mi = str.split("\\s");

        int[] flags = new int[q];

        for(int i=0;i<q;i++){
            int num = Integer.parseInt(mi[i]);
            for(int j=0;j<n;j++){
                int temp1 = Integer.parseInt(A[j]);
                int sum = temp1;
                if(num == temp1){
                    flags[i]+=1;
                }
                for(int k=j+1;k<n;k++){
                    int temp2 = Integer.parseInt(A[k]);
                    sum += temp2;
                    if(num == (temp1+temp2)){
                        flags[i]+=1;
                    }else if(num == sum){
                        flags[i]+=1;
                    }
                }
            }
        }

        for(int i=0;i<q;i++){
            if(flags[i] > 0){
                System.out.println("yes");
            }else{
                System.out.println("no");
            }
        }
        

    }    
}