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: Runtime Error)

#include <cstdio>
#include <vector>
#include <algorithm>

using namespace std;

bool can_make(const vector<int> A, int m, vector<vector<int>> dp, size_t i=0) {
    if (i == A.size()) return (dp[m][i] = (m==0));
    if (!m) return (dp[m][i] = 1);
    if (m < 0) return false;
    if (~dp[m][i]) return dp[m][i];

    if (m >= A[i])
        if (can_make(A, m-A[i], dp, i+1))
            return (dp[m][i] = 1);

    return (dp[m][i] = can_make(A, m, dp, i+1));
}

int main() {
    size_t n;
    scanf("%zu", &n);

    vector<int> A(n);
    for (size_t i=0; i<n; ++i)
        scanf("%d", &A[i]);

    size_t q;
    scanf("%zu", &q);

    for (size_t i=0; i<q; ++i) {
        int m;
        scanf("%d", &m);

        vector<vector<int>> dp(m+1, vector<int>(A.size(), -1));
        printf("%s\n", can_make(A, m, dp)? "yes":"no");
    }

    return 0;
}