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>
#include <stdlib.h>

int sums(int n, int ans, int *A);

int main()
{
    int n,m,i,con;
    int A[2000] = {0};
    int B[2000] = {0};

    scanf("%d",&n);

    for (i=0; i<n; i++) {
        scanf("%d",&A[i]);
    }
    
    scanf("%d",&m);

    for (i=0; i<m; i++) {
        scanf("%d",&B[i]);
    }

    for (i=0; i<m; i++) {
        con = sums(n,B[i],A);
        if (con) printf("yes");
        else printf("no");

        printf("\n");
    }

    return 0;
}

int sums(int n, int ans, int *A) {
    int t,s;

    if (n == 0) {
        if (ans == 0) return 1;
        else return 0;
    }

    //if (A[n-1] > ans) return 0;

    if (A[n-1] == ans) return 1;

    t = sums(n-1,ans,A);
    s = sums(n-1,ans-A[n-1],A);

    return (t || s);
}