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)

//?度?n 的数列a 中任意几个元素相加能否得到 k,能得到?出yes  ，否??出no

#include <iostream>
#include <cstdio>
using namespace std;
int n,m;
int a[10000];
int solve(int i,int p)
{
    if(p == 0)
        return 1;
    if(i >= n)   //没有找到
        return 0;
    int res = solve(i+1,p)||solve(i+1,p-a[i]);
    return res;
}
int main()
{
    int k;
    scanf("%d",&n);
    for(int i=0;i<n;i++)
        scanf("%d",&a[i]);
    scanf("%d",&m);
    for(int j=0;j<m;j++)
    {
        scanf("%d",&k);
        if(solve(0,k))
            printf("yes\n");
        else
            printf("no\n");
    }
    return 0;
}