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,q;
int c[20];
 int sum(int a,int m,int i);
  int max(int v,int y);
int main()
{

	int b[200];
	scanf("%d",&n);
	for(int i = 0; i < n; i++)
		scanf("%d",&c[i]);
	scanf("%d",&q);
	for(int i = 0; i < q; i++)
		scanf("%d",&b[i]);
	for(int i = 0; i < q; i++)
	{
		int v;
		v = sum(b[i],0,0);
		if(v == 1)printf("yes\n");
		else printf("no\n");
    } 
}
 
 int sum(int a,int m,int i)
 {
 	if(m==a)	return  1;
 	else if(m<a&&i<=n-1){
			return max(sum(a,m,i+1),sum(a,m+c[i],i+1));
	 }
	 else return 0;
 }
 
 int max(int v,int y)
 {
 	if (v>y) return v;
 	else return y;
 }