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 <iostream>
#include <cmath>
#include <vector>
#include <string>
#include <algorithm>

using namespace std;

#define INF 1e9
#define MAXN 100005
#define MAXM 100005
#define MOD 1000000007
#define ll long long
#define vi vector<int>
#define vll vector<long long>
#define rep(i,n) for(int i=0, i##_len=(n); i<i##_len; ++i)
#define pii pair<int, int>

ll ans;
int N,Q;
vi A, M;

bool canMake(int id, int m){
    if(m == 0) return true;
    if(id >= N) return false;

    bool res = canMake(id+1, m - A[id]) || canMake(id+1, m);
    return res;
}

void solve(){
    rep(i, Q){
        if(canMake(0, M[i])) cout << "yes" << endl;
        else cout << "no" << endl;
    }
}

int main(){
    cin >> N;
    A.resize(N);
    rep(i, N) cin >> A[i];

    cin >> Q;
    M.resize(Q);
    rep(i, Q) cin >> M[i];

    solve();
}