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 JavaScript (judge verdict: Accepted)

(function main() {
  const lines = require('fs').readFileSync('/dev/stdin', 'utf8').trim().split('\n');
  const n1 = Number(lines.shift());
  const arr1 = lines.shift().split(' ').map(Number);
  const n2 = Number(lines.shift());
  const arr2 = lines.shift().split(' ').map(Number);

  const solve = memoization(function(i, m) {
    if (m == 0) return true;
    if (i >= n1) return false;
    return solve(i + 1, m) || solve(i + 1, m - arr1[i]);
  });

  arr2.map(m => solve(0, m) ? 'yes' : 'no').forEach(res => console.log(res));
})();

function memoization(f) {
  let cache = {};
  return function innerFunc(...param) {
    const c = cache[param];
    return typeof c === 'undefined' ? cache[param] = f(...param) : c;
  }
}