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>
#include<vector>

using namespace std;

int n, q;
vector<int> A(n);

bool solve( const int p, int t )
{
	if( 0 == t ) { return true; }
	if( p < n ) {
		return solve( p+1, t - A.at( p ) ) | solve( p+1, t );
	} else {
		return false;
	}
}

int main()
{
	cin>>n;
	int num;
	for( int i=0; i<n; i++ ) {
		cin>>num;
		A.push_back( num );
	}
	cin>>q;
	for( int i=0; i<q; i++ ) {
		cin>>num;
		if( solve( 0, num ) ) {
			cout<<"yes"<<'\n';			
		} else {
			cout<<"no"<<'\n';
		}
	}
    return 0;
}