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: Time Limit Exceeded)

#include <stdio.h>

void del_element(int *a,int i,int n, int* temp){
    int j,k=0;
    for(j=0;j<n;j++){
        if(j!=i){
            temp[k]=a[j];
            k++;
        }
    }
}

int solve(int *a,int n,int m){
    if (m == 0) return 1;
    if (m < 0) return 0;
    int i;
    int calc=0;
    int temp[21];

    for(i=0;i<n;i++){
        calc=a[i];
        if(calc==m){
            return 1;
        }
        else{
            del_element(a,i,n,temp);
            if(solve(temp,n-1,m-calc)==1){
                return 1;
            }
        }
    }
    return 0;
}

int main(){
    int n,q,m[201];
    int A[21];
    scanf("%d",&n);
    int i;
    for(i=0;i<n;i++){
        scanf("%d",&A[i]);
    }
    scanf("%d",&q);
    for(i=0;i<q;i++){
        scanf("%d",&m[i]);
    }
    for(i=0;i<q;i++){
        if(solve(A,n,m[i])==1){
            printf("yes\n");
        }
        else{
            printf("no\n");
        }
    }
    return 0;
}