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++14 (judge verdict: Accepted)

#include<iostream>
#include<vector>
#include<string>
#include<algorithm>
#include<map>
#include<queue>
#include<stack>
#include<cmath>	
#include<list>

using namespace std;

bool solve(vector<int>& A, int n, int i, int m) {
	if (m == 0)return 1;
	if (i >= n)return 0;

	int res = solve(A, n, i + 1, m) || solve(A, n, i + 1, m - A[i]);

	return res;
}

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

	}

	return 0;
}