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: Accepted)

import java.util.*;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt(); //数列Aの長さ
        int[] A = new int[n]; //数列A
        for (int i=0; i<n; i++) {
            A[i] = sc.nextInt();
        }
        int q = sc.nextInt(); //整数mの数
        for (int i=0; i<q; i++) {
            int m = sc.nextInt(); //整数m
            if(check(n,m,A)) {
                System.out.println("yes");
            } else {
                System.out.println("no");
            }            
        }
    }
    static boolean check(int n, int m, int[] A) {
        if(m==0) return true;
        else if(n==0) return false;
        else if(check(n-1,m,A)) return true;
        else if(check(n-1,m-A[n-1],A)) return true;
        else return false;
    }
}