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<iostream>
using namespace std;

int n,q;
int A[21];
bool dp[21][2000];

bool solve(int i,int k){
    if(dp[i][k]==true) return true;
    if(k==0)
        dp[i][k]=true;
    else if(i>=n){
        dp[i][k]=false;
    }else if(solve(i+1,k)){
        dp[i][k]=true;
    }else if(solve(i+1,k-A[i])){
        dp[i][k]=true;
    }else{
        dp[i][k]=false;
    }

    return dp[i][k];

}
void deal(int k){
    for(int i=0;i<n;i++) {
        for(int j=1;j<2000;j++)
            dp[i][j]=false;
    }
    if(solve(0,k)){
        cout<<"yes"<<endl;
    }else{
        cout<<"no"<<endl;
    }
}

int main(){
    int k;
    cin>>n;
    for(int i=0;i<n;i++){
        cin>>A[i];
        dp[i][0]=true;
    }
    cin>>q;
    for(int i=0;i<q;i++){
        cin>>k;
        deal(k);
    }
    return 0;
}