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 Rust (judge verdict: Accepted)

use std::io::Read;

fn main() {
    let mut buf = String::new();
    std::io::stdin().read_to_string(&mut buf).unwrap();

    let answer = solve(&buf);

    println!("{}", answer);
}

fn solve(input: &str) -> String {
    let mut iterator = input.split_whitespace();

    let n: usize = iterator.next().unwrap().parse().unwrap();
    let a: Vec<usize> = (0..n).map(|_| iterator.next().unwrap().parse().unwrap()).collect();
    let q: usize = iterator.next().unwrap().parse().unwrap();

    let mut candidates = std::collections::HashSet::new();
    for i in 1..2_u64.pow(n as u32) {
        let bs: Vec<char> = binary_string(i as usize, n).chars().collect();
        let mut sum = 0;
        for j in 0..n {
            if bs[j] == '1' {
                sum += a[j];
            }
        }
        candidates.insert(sum);
    }

    let mut result = String::new();
    for _ in 0..q {
        let m: usize = iterator.next().unwrap().parse().unwrap();

        if candidates.contains(&m) {
            result.push_str("yes\n");
        } else {
            result.push_str("no\n");
        }
    }

    return result.trim().to_string();
}

fn binary_string(n: usize, keta: usize) -> String {
    return format!("{:0>1$b}", n, keta);
}