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++14 (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
const ll INF = 1LL << 60;
const int MAX = 1000000010;
const ll MOD = 1000000007;

int main() {
    int N; cin >> N;
    vector<int> A(N);
    for (int i = 0; i < N; i++) {
        cin >> A[i];
    }
    
    int q; cin >> q;
    vector<int> B(q);
    vector<bool> C(q, false);
    for (int i = 0; i < q; i++) cin >> B[i];

    for (int tmp = 0; tmp < (1 << N); tmp++) {
        bitset<20> b(tmp);
        int sum = 0;

        for (int i = 0; i < N; i++) {
            if (b.test(i)) {
                sum += A[i];
            }
        }

        for (int i = 0; i < q; i++) {
            if (sum == B[i]) {
                C[i] = true;
            }
        }
    }

    for (int i = 0; i < q; i++) {
        if (C[i]) {
            cout << "yes" << endl;
        } else {
            cout << "no" << endl;
        }
    }
    return 0;
}