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 <string>
#include <vector>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <cstdlib>

#define rep(i, n)     for (int (i) = 0; (i) < (n); (i)++)
#define rep1(i, n)    for (int i = 1; i <= (int)(n); i++)

using namespace std;
typedef long long ll;

int n, A[50];

int solve(int i, int m) {
   if (m == 0) return(true);
   else if (i >= n) return(false);

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

int main() {
   int q, m;

   cin >> n;
   rep(i, n) cin >> A[i];
   cin >> q;
   rep(i, q) {
      cin >> m;
      if (solve(0, m)) cout << "yes" << endl;
      else cout << "no" << endl;
   }
   return(0);
}