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)

let inputString = '';

process.stdin.on('data', (input) => { inputString += input; });

process.stdin.on('end', () => {
  const arr = inputString.split('\n');
  check(arr[1].split(' ').map((s) => Number(s)), arr[3].split(' ').map((s) => Number(s)));
});

const check = (A, mList) => {
  const solve = (i, m) => {
    if (m === 0) return true;

    if (i >= A.length) return false;

    // solve(i + 1, m) is the case which does not use A[i] in the addition
    // solve(i + 1, m - A[i]) is the case which uses A[i] in the addition
    return solve(i + 1, m) || solve(i + 1, m - A[i]);
  };

  mList.forEach((m) => {
    console.log(solve(0, m) ? 'yes' : 'no');
  });
};