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 <math.h>

int judge(int *L, int idx, int m){
    int i;
    for (i = 0; i < idx; i++)
    {
        if (L[i] == m) return 1;
    }
    return 0;
}

int main(void){
    // Your code here!
    int i, j, n, A[20], q, m, L[1500000], idx;
    long x, temp;
    double xx;
    scanf("%d", &n);
    for (i = 0; i < n; i++) scanf("%d", &A[i]);
    // x = 2 ^ n;
    xx = pow(2.0, (double)n);
    x = (long)xx;
    idx = 0;
    for (i = 0; i < x; i++)
    {
        temp = 0;
        for (j = 0; j < n; j++)
        {
            if (i & (1 << j)) temp += A[j];
        }
        L[idx] = temp;
        L[idx + 1] = -1;
        idx++;
    }
    /*
    printf("x = %d\n", x);
    for (i = 0; i < idx; i++) printf("%d ", L[i]);
    printf("\n");
    */
    scanf("%d", &q);
    for (i = 0; i < q; i++)
    {
        scanf("%d", &m);
        if (judge(L, idx, m) == 1) printf("yes\n");
        else printf("no\n");
    }
    return 0;
}