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 solve( int ,int ) ;
int n, a[20], ans = 0 ;
int main () {
	
	int m[200], i, j, q;
	int count = 0 ;

	
	scanf ("%d",&n) ;
	for ( i = 0 ; i < n ; i ++ ) {
		scanf("%d", &a[i] ) ;
	}
	scanf ("%d",&q) ;
	for ( j = 0 ; j < q ; j ++ ) {
		scanf("%d", &m[j] ) ;
	}
	
	
	for ( j = 0 ; j < q ; j++) {
		ans = 0 ;
		solve ( 0 , m[j] );
		if ( ans == 1 ) {
			printf("yes\n");
		}
		else { 
			printf("no\n");
		}
	}
	
return 0 ;
}


int solve( int j ,int m ) {
	if (m == 0 ) {
		ans =  1;
		return 1 ; 
	}
	
	else if ( j < n ) {
		solve( j + 1 , m) ;
		solve( j + 1 , m - a[j]) ;
	}
}