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>
#define MAX 20
#define MAX2 200

int keisan(int,int);
int a[MAX],b[MAX2],count=0,sum,n;
int main() {
int h,i,j;
int m,key,v;
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]);
  v=keisan(0,b[i]);
     if(v==1) printf("yes\n");
     else printf("no\n");
}
  return 0;
}
  int keisan(int x,int y){
    if(y == 0) return 1;
      if(x >= n) return 0;
      return keisan(x + 1,y) || keisan(x + 1,y - a[x]);
  }