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 add_A(int *,int,int);//引数 （配列、使用できる数字の最大があるインデックス,qの現在値）
int main(){
  int i/*ループ用*/ ,j/*ループ用*/ ,A[20] ,n ,q ,m ,v/*ソート用数字避難*/;
  
  scanf("%d",&n);
  for(i=0;i<n;i++)scanf("%d",&A[i]);

  for(i = 1; i<n;i++){//使用できる数字を昇順にソート
    v = A[i];
    j = i - 1;
    while( j >= 0 && A[j] > v){
      A[j+1] = A[j];
      j--;
    }
    A[j+1] = v;
  }

  scanf("%d",&q);
  for(j=0;j<q;j++){
    scanf("%d",&m);
    if(add_A(A,n-1,m))printf("yes\n");//関数の戻値によって表示を変化
    else printf("no\n");
  }
  return 0;
}
int add_A(int *array,int n,int m){
  int i;
  
  for(i=n;i>=0;i--){
    
    if(m == array[i])return 1;
    if(m - array[i]>0){
      if(add_A(array,i-1 ,m - array[i]))return 1;
    }
  }
  return  0;
}