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>

/*A[i]をmから引いていき0を作れるか全検索する*/
int Solve(int i , int m , int A[] , int n) {
    int respond;
    /*配列Aいくつかの組み合わせで引いていって、mを0にすることができたら*/
    if(m == 0)
        return 1;
    /*iを増やしていった結果、配列Aの要素数よりも大きくなってしまった ＝ 足した結果、それでもmに届かなかった*/
    /*C言語で偽は0それ以外は真*/
    if(i >= n)
        return 0;
    
    /*ここからが再帰(再帰をするときは必ず無限ループしないようにすることに注意)*/
    /*||はorなので「右が偽かつ左も偽なら偽」「右か左のどちらか一方が真ならばもう片方の真偽によらず真」*/
    /*ここではSolve(i ＋ 1 , m）と(i + 1 , m - A[i])の両方が常に実行されるので2^n回行われる*/
    /*mからA[i]を引かない場合と、引く場合の二通りがある*/
    respond = Solve(i + 1 , m , A , n) || Solve(i + 1 , m - A[i] , A , n);
    
    /*res = ~は再帰なので、resが真か偽かの判定を出すまで終わらない*/
    /*結果としてresが返した答えを返す(真or偽の二択)*/
    return respond;
}

int main(void)
{
    int n , q , m;
    int i , j;
    int count = 0;
    
    scanf("%d" , &n);
    
    int A[n];
    for(i = 0 ; i < n ; i++)
        scanf("%d" , &A[i]);
    
    scanf("%d" , &q);
    
    for(j = 0 ; j < q ; j ++) {
        scanf("%d" , &m);
        /*再帰なので0から始めて全て調べる*/
        if(Solve(0 , m , A , n) == 1)
            puts("yes");
        else
            puts("no");
    }

    return 0;
}