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 boolean[] vis;                       //line 74 and 91 hasElement ans pth_fnd
    static int MOD=1000000007;
    //
    static int[] ans;
    static int cnt;
    static long set1=0,set2=0;
    //
    public static void main(String[] args) throws IOException{
        // long sttm=System.currentTimeMillis();
        FastReader sc=new FastReader();
        OutputStream out=new BufferedOutputStream(System.out);
        int t=sc.nextInt();
        while(t-->0){
            int n=sc.nextInt(),k=sc.nextInt();
            int[] a=new int[n];
            HashSet<Integer> set=new HashSet<Integer>();
            for(int i=0;i<n;i++){
                a[i]=sc.nextInt();
                set.add(a[i]);
            }
            int alpha=-1;
            boolean con=true;
            if(set.size()>k){
                out.write(("-1\n").getBytes());
            }
            else{
                ArrayList<Integer> arrl=new ArrayList<Integer>(set);
                while(arrl.size()<k){
                    arrl.add(arrl.get(0));
                }
                out.write((arrl.size()*n+"\n").getBytes());
                for(int i=0;i<n;i++){
                    for(int j=0;j<arrl.size();j++){
                        out.write((arrl.get(j)+" ").getBytes());
                    }
                }
                // System.out.println();
                out.write(("\n").getBytes());
            }
        }
        out.flush();
    }


    public static boolean dfsCycle(ArrayList<Integer>[] arrl,int[] vis,int st,int pr){
        vis[st]=1;
        int cupr=st;
        for(int elm:arrl[st]){
            if(vis[elm]==1 && elm!=pr){
                return true;
            }
            else if(vis[elm]==0 && dfsCycle(arrl, vis, elm, cupr)){
                return true;
            }
        }
        vis[st]=2;
        return false;
    }
    


    public static long powMOD(long n,long k){                 //n-pow(k)
        if(k==1){
            return n%MOD;
        }
        if(k==0){
            return 1;
        }
        long val=pow(n,k/2)%MOD;
        if(k%2==0)return ((val%MOD)*(val%MOD))%MOD;
        else return (((val*val)%MOD)*n)%MOD;
    }

    public static long pow(long n,long k){                 //n-pow(k)
        if(k==1){
            return n;
        }
        if(k==0){
            return 1;
        }
        long val=pow(n,k/2);
        if(k%2==0)return ((val)*(val));
        else return (((val*val))*n);
    }


    public static HashMap<Integer,Integer> lcm(int num){
        ArrayList<Integer> arrl=new ArrayList<Integer>();
        HashMap<Integer,Integer> map=new HashMap<Integer,Integer>();
        while(num%2==0){
            
            if(!map.containsKey(2)){
                arrl.add(2);
                map.put(2, 1);
            }
            else{
                map.put(2,map.get(2)+1);
            }
            num/=2;
        }
        for(int i=3;i*i<=num;i+=2){
            while(num%i==0){
                if(!map.containsKey(i)){
                    arrl.add(i);
                    map.put(i, 1);
                }
                else{
                    map.put(i,map.get(i)+1);
                }
                num/=i;
            }
        }
        if(num>2){
            if(!map.containsKey(num)){
                arrl.add(num);
                map.put(num, 1);
            }
            else{
                map.put(num,map.get(num)+1);
            }
        }
        return map;
    }


    static long m=998244353l;
    static long modInverse(long a,long m) 
    { 
        long m0 = m;
        long y = 0l, x = 1l; 
  
        if (m == 1) 
            return 0l; 
  
        while (a > 1) 
        { 
  
            long q = a / m; 
  
            long t = m; 
  
            m = a % m; 
            a = t; 
            t = y; 
  
            y = x - q * y; 
            x = t; 
        } 
        if (x < 0) 
            x += m0; 
  
        return x; 
    }


    public static int num_fact(int num){
        ArrayList<Integer> arr=new ArrayList<Integer>();
        int cnt=0;
        for(int i=1;i*i<=num;i++){
            if(num%i==0){
                if(i*i==num){
                    cnt+=1;
                }
                else{
                    cnt+=2;
                }
            }
        }
        return cnt;
    }


    public static void sort_inc(int[][] arr,int col){        //change dimention if req
        Arrays.sort(arr, new Comparator<int[]>() {
            public int compare(final int[] entry1,final int[] entry2) { 
                if (entry1[col] > entry2[col])              //this is for inc
                    return 1; 
                else if(entry1[col]==entry2[col])
                    return 0;
                else
                    return -1; 
            }
        });
    }


    public static boolean prime(int n){
        for(int i=2;i*i<=n;i++){
            if(n%i==0){
                return false;
            }
        }
        return true;
    }


    static long gcd(long a, long b) 
	{ 
	if (b == 0) 
		return a; 
	return gcd(b, a % b); 
    } 
    
    public static long lcmOfTwo(int a,int b){
        return a*b/gcd(a,b);
    }

}


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; 
    }
}