Problem: Phoenix loves beautiful arrays. An array is beautiful if all its subarrays of length k have the same sum. A subarray of an array is any sequence of consecutive elements.

Phoenix currently has an array a of length n. He wants to insert some number of integers, possibly zero, into his array such that it becomes beautiful. The inserted integers must be between 1 and n inclusive. Integers may be inserted anywhere (even before the first or after the last element), and he is not trying to minimize the number of inserted integers.

Input

The input consists of multiple test cases. The first line contains an integer t (1 ≤ t ≤ 50) — the number of test cases.

The first line of each test case contains two integers n and k (1 ≤ k ≤ n ≤ 100).

The second line of each test case contains n space-separated integers (1 ≤ a_i ≤ n) — the array that Phoenix currently has. This array may or may not be already beautiful.

Output

For each test case, if it is impossible to create a beautiful array, print -1. Otherwise, print two lines.

The first line should contain the length of the beautiful array m (n ≤ m ≤ 10^4). You don't need to minimize m.

The second line should contain m space-separated integers (1 ≤ b_i ≤ n) — a beautiful array that Phoenix can obtain after inserting some, possibly zero, integers into his array a. You may print integers that weren't originally in array a.

If there are multiple solutions, print any. It's guaranteed that if we can make array a beautiful, we can always make it with resulting length no more than 10^4.

Example

Input


4
4 2
1 2 2 1
4 3
1 2 2 1
3 2
1 2 3
4 4
4 3 4 2


Output


5
1 2 1 2 1
4
1 2 2 1
-1
7
4 3 2 1 4 3 2

Note

In the first test case, we can make array a beautiful by inserting the integer 1 at index 3 (in between the two existing 2s). Now, all subarrays of length k=2 have the same sum 3. There exists many other possible solutions, for example: 

  * 2, 1, 2, 1, 2, 1 
  * 1, 2, 1, 2, 1, 2 



In the second test case, the array is already beautiful: all subarrays of length k=3 have the same sum 5.

In the third test case, it can be shown that we cannot insert numbers to make array a beautiful.

In the fourth test case, the array b shown is beautiful and all subarrays of length k=4 have the same sum 10. There exist other solutions also.

Solution (in JAVA):

Solution:

import java.util.*;
import java.io.*;
public class Main {
    static FastReader in=new FastReader();
    static StringBuilder Sd=new StringBuilder();
    static List<Integer>Gr[]; 
    static int a[],b[];
    static boolean visit[];
    public static void main(String [] args) {
    	//Dir by MohammedElkady
    	int t=in.nextInt();
    	while(t-->0) {
    	int n=in.nextInt(),k=in.nextInt();
        boolean visit[]=new boolean[n+10],non=false;
        int a[]=fillint(n);
        int size=0;
        for(int i=0;i<k;i++) {
        	if(visit[a[i]]) {size++;}
        	visit[a[i]]=true;}
        for(int i=0;i<n;i++) {if(!visit[a[i]]) {non=true;
        visit[a[i]]=true;
        size--;}}
        ArrayList <Integer>let=new ArrayList();
        if(size>=0) {
        	for(int i=0;i<=n;i++) {
        		if(visit[i])let.add(i);
        	}
        	while(let.size()<k) {
        		let.add(let.get(0));
        	}
        	StringBuilder make=new StringBuilder();
        	int m=0;
        	for(int i=0,u=0;i<n;u++) {m++;
        	make.append(let.get(u%let.size())+" ");
        	if(let.get(u%let.size())==a[i]) {i++;}
        	}
        	Soutln(m+"");
        	Soutln(make.toString());
        }
        else if(non) {
        	Soutln("-1");
        }
        else {
        	StringBuilder x=new StringBuilder();
        	int m=n,l=0;
        	for(int i=0,u=0;i<n;) {
        		if(i<k) {x.append(a[i]+" ");i++;}
        		else {
        			x.append(a[u%k]+" ");
        			if(a[i]==a[u%k]) {i++;}
        			else m++;
        			u++;
        		}
        	}
        Soutln(m+"");
        Soutln(x.toString());
        }
    	}
    	
    Sclose();
    
    }
    static long gcd(long g,long x){
        if(x<1)return g;
        else return gcd(x,g%x);
    }
    //array fill
    static long[]filllong(int n){long a[]=new long[n];for(int i=0;i<n;i++)a[i]=in.nextLong();return a;}
    static int[]fillint(int n){int a[]=new int[n];for(int i=0;i<n;i++)a[i]=in.nextInt();return a;}
    //OutPut Line
    static void Sout(String S) {Sd.append(S);}
    static void Soutln(String S) {Sd.append(S+"\n");}
    static void Soutf(String S) {Sd.insert(0, S);}
    static void Sclose() {System.out.println(Sd);}
    static void Sclean() {Sd=new StringBuilder();}
}
 
class node implements Comparable<node>{
int x,p;
	node(int x,int p){
		this.x=x;
		this.p=p;
	}
	@Override
	public int compareTo(node o) {
       return (x-p)-(o.x-o.p);
	}
}
class FastReader
    {
        BufferedReader br;
        StringTokenizer st;
 
        public FastReader()
        {
            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;
        }
}
 
class Sorting{
 
 public static int[] bucketSort(int[] array, int bucketCount) {
 
 if (bucketCount <= 0) throw new IllegalArgumentException("Invalid bucket count");
 
 if (array.length <= 1) return array; //trivially sorted
 
 
 
 int high = array[0];
 
 int low = array[0];
 
 for (int i = 1; i < array.length; i++) { //find the range of input elements
 
 if (array[i] > high) high = array[i];
 
 if (array[i] < low) low = array[i];
 
 }
 
 double interval = ((double)(high - low + 1))/bucketCount; //range of one bucket
 
 
 ArrayList<Integer> buckets[] = new ArrayList[bucketCount];
 
 for (int i = 0; i < bucketCount; i++) { //initialize buckets
 
 buckets[i] = new ArrayList();
 
 }
 
 
 
 for (int i = 0; i < array.length; i++) { //partition the input array
 
 buckets[(int)((array[i] - low)/interval)].add(array[i]);
 
 }
 
 
 int pointer = 0;
 
 for (int i = 0; i < buckets.length; i++) {
 
 Collections.sort(buckets[i]); //mergeSort
 
 for (int j = 0; j < buckets[i].size(); j++) { //merge the buckets
 
 array[pointer] = buckets[i].get(j);
 
 pointer++;
 
 }
 
}
 
return array;
 
}
}