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>

int solve(int,int);

int q,n,A[2000],M[2000];

int main(){
  int i,j;

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

  //M[1~q]まで再帰にかける&out                                                                          
  for(i=0; i<q; i++){
    if( solve(0, M[i]) ){
      printf("yes\n");
    }
    else{
      printf("no\n");
    }
  }
  return 0;
}

int solve(int i,int m){
  //(M[]の内1項)−(分岐点の数値)=0ならその数値になり得る                                             
  if(m-A[i]==0) return 1;
  //A[]の要素数を超えるとmになりえない                                                          
  if(i >= n) return 0;
  //まだ判断ができない場合はsolveに再帰的に変数値を引き継ぐ                               
  if(solve(i+1,m-A[i]) || solve(i+1,m)) return 1;
  return 0;
}