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<stdlib.h>
#include<iostream>
#include<bits/stdc++.h>
#include<string.h>
using namespace std;
const int N = 2e5+10;
int a[N];
int n;
int solve(int i , int m)
{
	if(m == 0)
	  return 1;
	if(i >= n)
	  return 0;
	int res = solve(i+1,m)||solve(i+1,m-a[i]);
	return res;
}
int main()
{
	int q , M;
	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))
		  printf("yes\n");
		else
		  cout << "no" << endl;
	}
	return 0;
}