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<stdlib.h>
int exsearch(int*, int, int);
main()
{
        int i, n, q, m, *a;
        scanf("%d", &n);
        a = (int*)malloc(sizeof(int) * n);
        for(i = 0; i < n; i++){
        scanf("%d%*c", &a[i]);
        }
        scanf("%d", &q);
        for(i = 0; i < q; i++){
                scanf("%d%*c", &m);
                if(exsearch(a, m, n)){
                        printf("yes\n");
                }else{
                        printf("no\n");
                }
        }
        free(a);
}
int exsearch(int *a, int m, int n)
{
        int result = 0;
        if(!m){
        return 1;
        }else if(!n || m < 0 ){
                return 0;
        }else{
                result = exsearch(a, m - a[n - 1], n - 1) || exsearch(a, m, n - 1);
        }
        return result;
}