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 C (judge verdict: Accepted)

#include<stdio.h>

int n;
int A[2000];

int solve(int i, int m){

	int res;
	
	if( m == 0 ){
		return 1;
	}else if( i >= n ){
		return 0;
	}
	
	res = solve(i + 1, m) || solve(i + 1, m - A[i]);
	return res;
}
	
int main( void ){
	
	int q;
	int i;
	int m[2000];
	
	scanf( "%d" , &n );
	
	for( i = 0; i < n; i++ ){
		scanf( "%d" , &A[i] );
	}
	
	scanf( "%d" , &q );
	
	for( i = 0; i < q; i++ ){
		scanf( "%d" , &m[i] );
	}
	
	for( i = 0; i < q; i++ ){
		if( solve( 0 , m[i] ) == 1 ){
			printf( "yes\n" );
		}else{
			printf( "no\n" );
		}
	}
	
}