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: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>
void YesOrNo( int * combination, int * array_a , int * array_m, int * ans, int number_a, int number_m )
{
        int sum = 0;
        int i = 0, j = 0;
        for ( i = 0 ; i < number_m ; i++ )
        {
                sum = 0;
                for ( j = 0 ; j < number_a ; j++ )
                {
                        sum = sum + combination[j] * array_a[j];
                }

                if ( sum == array_m[i] )
                {
                        ans[i] = ans[i] + 1;
                        break;
                }
        }
}


void Recycle( int * combination, int * array_a, int * array_m, int * ans, int status, int number_a, int number_m )
{
        if ( status < 0 )
        {
                YesOrNo( combination, array_a, array_m, ans, number_a, number_m );
        }
        else
        {
                combination[status] = 1;
                Recycle( combination, array_a, array_m, ans, status - 1, number_a, number_m );
                combination[status] = 0;
                Recycle( combination, array_a, array_m, ans, status - 1, number_a, number_m );
        }
}

int main()
{
        int number_a, number_m;
        scanf( "%d", &number_a );
        int array_a[number_a];
        int combination[number_a];
        int i = 0;
        for ( i = 0 ; i < number_a ; i++ )
        {
                scanf( "%d", &array_a[i] );
        }

        scanf( "%d", &number_m );
        int array_m[number_m];
        int ans[number_m];
        for ( i = 0 ; i < number_m ; i++ )
        {
                scanf( "%d", &array_m[i] );
                ans[i] = 0;
        }

        Recycle( combination, array_a, array_m, ans, number_a, number_a, number_m );
        for ( i = 0 ; i < number_m ; i++ )
        {
                if ( ans[i] > 0 )
                {
                        fprintf( stdout, "yes\n" );
                }
                else
                {
                        fprintf( stdout, "no\n" );
                }
        }
        return 0;
}