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>

/* 数列Aの長さ */
static int n;
/* 数列Aへのポインタ */
static int *A;

/* 入力値の M から選んだ要素を引いていく再帰関数 */
/* 再帰呼出されると i をカウントアップする */
int solve(int i, int m)
{
    if (m == 0)
        return (1); /* 0 の場合は"yes"を返すことにする */
    if (i >= n)
        return (0);

    int res = solve(i + 1, m) || solve(i + 1, m - A[i]);
    return (res);
}

int main()
{
    int q, M, i;

    /* 組み合わせに使用できる数値を入力 */
    scanf("%d", &n);
    A = (int *)calloc(n, sizeof(int));
    for (i = 0; i < n; i++) {
        scanf("%d", &A[i]);
    }
    /* 質問 */
    scanf("%d", &q);
    for (i = 0; i < q; i++) {
        scanf("%d", &M);
        if (solve(0, M)) {
            puts("yes");
        }
        else {
            puts("no");
        }
    }
    return (0);
}