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)

const input = require('fs').readFileSync('/dev/stdin', 'utf8');

const list = input.trim().split('\n');
const n = Number(list[0]);
const A = list[1].trim().split(' ');
const q = Number(list[2]);
const mi = list[3].trim().split(' ');

// const dp = {};
// const memo = (i, sum) => {
//   dp[sum] = true;
//   if (i === n) {
//     return;
//   }
//   memo(i + 1, sum);
//   memo(i + 1, sum + Number(A[i]));
// }
// memo(0, 0);

// for (let i = 0; i < q; i++) {
//   console.log(dp[mi[i]] ? 'yes' : 'no');
// }

const memo2 = (i, m) => {
  if (i === n) {
    return m === 0;
  }
  return memo2(i + 1, m) || memo2(i + 1, m - Number(A[i]));
}

for (let i = 0; i < q; i++) {
  console.log(memo2(0, Number(mi[i])) ? 'yes' : 'no');
}