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<bits/stdc++.h>
#define ll long long int
#define FOR(i,a,b) for(ll i=(ll)(a);i<(ll)(b);i++)
#define REP(i,n) FOR(i,0,n)
#define REP1(i,n) FOR(i,1,n)
using namespace std;

int A[50], 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;
  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;
}