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<iostream>
using namespace std;
const int MAXN=20005;
bool used[MAXN];
int main(){
	for(int i=0;i<MAXN;i++)used[i]=false;
	int n;
	int sum=1;
	cin>>n;
	for(int i=0;i<n;i++){
		int x;
		cin>>x;
		for(int j=0;j<=sum;j++){
			if(used[j])used[j+x]=true;
		}
		sum+=x;
		used[x]=true;
	}
	int q;
	cin>>q;
	for(int i=0;i<q;i++){
		int x;
		cin>>x;
		if(used[x])cout<<"yes"<<endl;
		else cout<<"no"<<endl;
	}
	return 0;
} */


#include<iostream>
using namespace std;
int n,A[50];
bool solve(int i,int m){
	if(m==0)return true;
	if(i>=n)return false;
	bool res=solve(i+1,m)||solve(i+1,m-A[i]);
	return res;
}

int main(){
	int q,M,i;
	cin>>n;
	for(int i=0;i<n;i++)cin>>A[i];
	cin>>q;
	for(int i=0;i<q;i++){
		cin>>M;
		if(solve(0,M))cout<<"yes"<<endl;
		else cout<<"no"<<endl;
	}
	return 0;
}