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(_stdin) {
    const stdin = _stdin.trim().split('\n');
    const n = Number(stdin.shift());
    const A = stdin.shift().split(' ').map(Number);
    const q = Number(stdin.shift());
    const m = stdin.shift().split(' ').map(Number);

    let totalResults = [];
    let max = '';
    for (let i = 1; i <= n; i++) {
        max += '1';
    }
    let combi = parseInt(max, 2);
    for (let i = 1, l = parseInt(combi, 10); i <= l; i++) {
        let op = i.toString(2);
        op = padding(op, n);
        const targets = A.filter((e, i) => op[i] === '1');
        totalResults.push(targets.reduce((t, c) => t += c));
    }

    for (let i = 0; i < q; i++) {
        console.log(totalResults.includes(m[i]) ? 'yes' : 'no');
    }
}

function padding(str, n) {
    if(str.length > n) return str;
    let diff = n - str.length;
    let pad = '';
    for(let i = 0;i<diff;i++){
        pad += '0';
    }
    return pad + str;
}

main(require('fs').readFileSync('/dev/stdin', 'utf8'));