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.*;
import java.io.*;

class Main {
	static boolean[] isExist = new boolean[2001];
	static int[] A;
	static int n;
    public static void main( String[] args ) {
        try {
	      	BufferedReader kb = new BufferedReader( new InputStreamReader( System.in ) );
			n = Integer.parseInt( kb.readLine() );
			String[] buf = kb.readLine( ).split( " " );
			A = new int[n];
			int sum = 0;
			for( int i=0; i<n; i++ ) {
				A[i] = Integer.parseInt( buf[i] );
				sum += A[i];
			}
			solve( 0, sum );
			
			StringBuilder output = new  StringBuilder( );
			int q = Integer.parseInt( kb.readLine() );
			buf = kb.readLine( ).split( " " );
			for( int i=0; i<q; i++ ) {
				if( isExist[ Integer.parseInt( buf[i] ) ] ) {
					output.append( "yes" ).append( "\n" );			
				} else {
					output.append( "no" ).append( "\n" );
				}
			}
			System.out.print( output );            
        } catch( IOException e ) {
            System.err.println( e );
        }
    }
	public static void solve( int p, int t )
	{
		isExist[t] = true;
		if( p < n ) {
			solve( p+1, t-A[p] );
			solve( p+1, t );
		}
		return;
	}
}