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 <stdio.h>
#include <stdbool.h>

int n_from;
int from[20];

int id;
bool canMake(int id, int target);

int main(void)
{
    int i;

    scanf("%d", &n_from);
    for (i = 0; i < n_from; i++) {
        scanf(" %d", &from[i]);
    }

    int n_target;
    scanf("%d", &n_target);
    int target;
    while (n_target--) {
        id = 0;
        scanf(" %d", &target);
        if (canMake(id, target)) {
            printf("yes\n");
        } else {
            printf("no\n");
        }
    }

    return 0;
}

bool canMake(int id, int target)
{
    /*
    int i;
    printf("judging if can make %d from", target);
    for (i = id; i < n_from; i++) {
        printf(" %d", from[i]);
    }
    printf("\n");
    */

    if (id == n_from) {
        return (target == 0);
    }

    if (canMake(id + 1, target)) {
        return true;
    }
    if (canMake(id + 1, target - from[id])) {
        return true;
    }

    return false;
}