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 C++ (judge verdict: Accepted)

#include<iostream>
#include<cstdio>
#include<queue>
using namespace std;

int countN, n[20], countQ, q;

bool solve(int i, int m) {
    if (m == 0) return true;
    if (i >= countN) return false;
    return solve(i + 1, m) || solve(i + 1, m - n[i]);
}

        
int main() {
    cin >> countN;
    for (int i = 0; i < countN; i++) {
        cin >> n[i];
    }

    cin >> countQ;
    for (int i = 0; i < countQ; i++) {
        cin >> q;
        if (solve(0, q)) printf("yes\n");
        else printf("no\n");
    }

    return 0;
}