Problem: Omkar has received a message from Anton saying "Your story for problem A is confusing. Just make a formal statement." Because of this, Omkar gives you an array a = [a_1, a_2, …, a_n] of n distinct integers. An array b = [b_1, b_2, …, b_k] is called nice if for any two distinct elements b_i, b_j of b, |b_i-b_j| appears in b at least once. In addition, all elements in b must be distinct. Can you add several (maybe, 0) integers to a to create a nice array b of size at most 300? If a is already nice, you don't have to add any elements.

For example, array [3, 6, 9] is nice, as |6-3|=|9-6| = 3, which appears in the array, and |9-3| = 6, which appears in the array, while array [4, 2, 0, 6, 9] is not nice, as |9-4| = 5 is not present in the array.

For integers x and y, |x-y| = x-y if x > y and |x-y| = y-x otherwise.

Input

Each test contains multiple test cases. The first line contains t (1 ≤ t ≤ 50), the number of test cases. Description of the test cases follows.

The first line of each test case contains a single integer n (2 ≤ n ≤ 100) — the length of the array a.

The second line of each test case contains n distinct integers a_1, a_2, ⋅⋅⋅, a_n (-100 ≤ a_i ≤ 100) — the elements of the array a.

Output

For each test case, output one line containing YES if Omkar can create a nice array b by adding elements to a and NO otherwise. The case of each letter does not matter, so yEs and nO will also be accepted.

If the first line is YES, output a second line containing a single integer k (n ≤ k ≤ 300). 

Then output one line containing k distinct integers b_1, b_2, ⋅⋅⋅, b_k (-10^9 ≤ b_i ≤ 10^9), the elements of the nice array b. b_1, b_2, ⋅⋅⋅, b_k can be in any order. For each a_i in a, a_i must appear at least once in b.

It can be proved that if Omkar can create such an array b, then he can also do so in a way that satisfies the above constraints.

If multiple solutions exist, you can print any. 

Example

Input


4
3
3 0 9
2
3 4
5
-7 3 13 -2 8
4
4 8 12 6


Output


yes
4
6 0 3 9
yEs
5
5 3 1 2 4
NO
Yes
6
8 12 6 2 4 10

Note

For the first case, you can add integers to a to receive the array b = [6, 0, 3, 9]. Note that |6-3| = |9-6| = |3-0| = 3 and 3 is in b, |6-0| = |9-3| = 6 and 6 is in b, and |9-0| = 9 is in b, so b is nice.

For the second case, you can add integers to a to receive the array b = [5, 3, 1, 2, 4]. We have that |2-1| = |3-2| = |4-3| = |5-4| = 1 is in b, |3-1| = |4-2| = |5-3| = 2 is in b, |4-1| = |5-2| = 3 is in b, and |5-1| = 4 is in b, so b is nice.

For the fourth case, you can add integers to a to receive the array b = [8, 12, 6, 2, 4, 10]. We have that |4-2| = |6-4| = |8-6| = |10-8| = |12-10| = 2 is in b, |6-2| = |8-4| = |10-6| = |12-8| = 4 is in b, |8-2| = |10-4| = |12-6| = 6 is in b, |10-2| = |12-4| = 8 is in b, and |12-2| = 10 is in b, so b is nice.

It can be proven that for all other test cases it is impossible to create a nice array b.

Solution (in JAVA):

Solution:
import java.util.*;
import java.io.*;
 
public class SolutionA{
       public static void main(String[] args) throws Exception{
               Fast sc=new Fast();
       StringBuilder sb=new StringBuilder();
        PrintWriter out=new PrintWriter(System.out);
       int t=sc.nextInt();
       while(t-->0){
        
         int n=sc.nextInt();
         int[] ar=new int[n];
          HashSet<Integer> set=new HashSet<>();
         for(int i=0;i<n;i++){ar[i]=sc.nextInt();
          set.add(ar[i]);
         }
         boolean ok=false;
         while(set.size()<=300){
          int[] br=new int[set.size()];
         int p=0;
         for(int a:set){
          br[p]=a;
          p++;
         }
         ok=false;
          for(int i=0;i<set.size();i++){
          for(int j=i+1;j<set.size();j++){
            int val=Math.abs(br[i]-br[j]);
            if(!set.contains(val)){
               ok=true;
            }
          }
         }
         if(ok==false)break;
         for(int i=0;i<br.length;i++){
          for(int j=i+1;j<br.length;j++){
            int val=Math.abs(br[i]-br[j]);
            if(!set.contains(val)){
               set.add(val);
            }
          }
         }
       }
         
         
         if(ok)out.println("NO");
         else{
            out.println("YES");
            out.println(set.size());
            for(int a:set)out.print(a+" ");
              out.println();
         }


        
       }
    out.close();
}

static void ReverSort(int[] ar){
    ArrayList<Integer> al=new ArrayList<>();
    for(int i=0;i<ar.length;i++){
        al.add(ar[i]);
    }

    Collections.sort(al);
      int j=0;
    for(int i=al.size()-1;i>=0;i--){

        ar[j]=al.get(i);
        j++;
    }
}
static void Sort(int[] ar){
    ArrayList<Integer> al=new ArrayList<>();
    for(int i=0;i<ar.length;i++){
        al.add(ar[i]);
    }

    Collections.sort(al);
      int j=0;
    for(int i=0;i<al.size();i++){

        ar[j]=al.get(i);
        j++;
     }
}
static int gcd(int a,int b){
    if(b==0) return a;
    else return  gcd(b,a%b);
}
}
class Pair{
    int x, y;
     Pair(int x,int y){
        this.x=x;
        this.y=y;
     }
}
class Fast{
  BufferedReader br; 
  StringTokenizer st;
  public Fast(){ br=new BufferedReader(new InputStreamReader(System.in)); }
  String next() 
        {  while (st == null || !st.hasMoreElements()) 
            {   try
                {     st = new StringTokenizer(br.readLine()); } 
                catch (IOException  e) 
                {    e.printStackTrace(); } 
            } 
            return st.nextToken(); 
        }
  int nextInt(){  return Integer.parseInt(next()); }
  long nextLong(){     return Long.parseLong(next()); } 
  double nextDouble() {     return Double.parseDouble(next()); } 
  String nextLine() 
        { 
            String str = ""; 
            try{      str = br.readLine(); } 
            catch (IOException e)  {  e.printStackTrace(); } 
            return str; 
        } 
}