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.lang.*;
import java.io.*;
import java.math.*;
public class Prac{     
    static class InputReader { 
        private final InputStream stream;
        private final byte[] buf = new byte[8192];
        private int curChar, snumChars;
        public InputReader(InputStream st) {
            this.stream = st;
        } 
        public int read() {
            if (snumChars == -1)
                throw new InputMismatchException();
            if (curChar >= snumChars) {
                curChar = 0;
                try {
                    snumChars = stream.read(buf);
                } 
                catch (IOException e) {
                    throw new InputMismatchException();
                }
                if (snumChars <= 0)
                    return -1;
            }
            return buf[curChar++];
        }
        public int ni() {
            int c = read();
            while (isSpaceChar(c)) {
                c = read();
            }
            int sgn = 1;
            if (c == '-') {
                sgn = -1;
                c = read();
            }
            int res = 0;
            do {
                res *= 10;
                res += c - '0';
                c = read();
            } while (!isSpaceChar(c));
            return res * sgn;
        } 
        public long nl() {
            int c = read();
            while (isSpaceChar(c)) {
            c = read();
            }
            int sgn = 1;
            if (c == '-') {
                sgn = -1;
                c = read();
            }
            long res = 0;
            do {
                res *= 10;
                res += c - '0';
                c = read();
            } while (!isSpaceChar(c));
            return res * sgn;
        } 
        public int[] nia(int n) {
            int a[] = new int[n];
            for (int i = 0; i < n; i++) {
                a[i] = ni();
            }
            return a;
        } 
        public String rs() {
            int c = read();
            while (isSpaceChar(c)) {
                c = read();
            }
            StringBuilder res = new StringBuilder();
            do {
                res.appendCodePoint(c);
                    c = read();
            } while (!isSpaceChar(c));
            return res.toString();
        }
        public String nextLine() {
            int c = read();
            while (isSpaceChar(c))
                c = read();
            StringBuilder res = new StringBuilder();
            do {
                res.appendCodePoint(c);
                c = read();
            } while (!isEndOfLine(c));
            return res.toString();
        } 
        public boolean isSpaceChar(int c) {
            return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
        }

        private boolean isEndOfLine(int c) {
            return c == '\n' || c == '\r' || c == -1;
        } 
    }
    public static class Key {

        private final int x;
        private final int y;

        public Key(int x, int y) {
            this.x = x;
            this.y = y;
        }

        @Override
        public boolean equals(Object o) {
            if (this == o) return true;
            if (!(o instanceof Key)) return false;
            Key key = (Key) o;
            return x == key.x && y == key.y;
        }

        @Override
        public int hashCode() {
            int result = x;
            result = 31 * result + y;
            return result;
        }

    }
    static class Pair{
        int x,y;
        public Pair(int x,int y){
            this.y=y;
            this.x=x;
        }
    }
   
   
    static PrintWriter w = new PrintWriter(System.out);
    static long mod=998244353L,mod1=1000000007;
       
    static boolean check(int arr[],int k){
        int sum=0;
        for(int i=0;i<k;i++)sum+=arr[i];
        int c=sum;
        for(int i=k;i<arr.length;i++){
            c+=arr[i];
            c-=arr[i-k];
            if(c!=sum)return false;
        }
        return true;
    }
    public static void main(String [] args){
        InputReader sc=new InputReader(System.in);
//        int t=sc.ni();
//        while(t-->0){
//            int n=sc.ni(),k=sc.ni();
//            char arr[]=sc.nextLine().toCharArray();
//            int c[]=new int[26];
//            for(int i=0;i<arr.length;i++)c[arr[i]-'a']++;
//            Arrays.sort(arr);
//            ArrayList<Character> ans=new ArrayList<>();
//            ans.add(arr[k-1]);
//            for(int i=k;i<n;i++){
//                if(c[arr[i]-'a']>=k+1){
//                    while(c[arr[i]-'a']>=k+1){
//                        ans.add(arr[i]);
//                        c[arr[i]-'a']-=(k+1);
//                    }
//                }
//            }
//            for(char c1:ans)w.print(c1);
//            w.println();
//            
//        }
        int t=sc.ni();
        while(t-->0){
            int n=sc.ni();
            int k=sc.ni();
            int a[]=new int[n];
            Set<Integer> set=new HashSet<>();
            for(int i=0;i<n;i++){
                a[i]=sc.ni();
                set.add(a[i]);
            }
            boolean c=check(a,k);
            if(c){
                w.println(n);
                for(int i=0;i<n;i++)w.print(a[i]+" ");
                w.println();
            }
            else{
                int arr[]=new int[10001];
                if(set.size()>k)w.println(-1);
                else{
                    Set<Integer> rem=new HashSet<>();
                    for(int i=1;i<=n;i++)rem.add(i);
                    
                    int j=0;
                    for(int i:set){
                        arr[j++]=i;
                        rem.remove(i);
                    }
                    for(int i:rem){
                        if(j==k)break;
                        arr[j++]=i;
                    }
                    while(j<k){
                        arr[j++]=1;
                    }
                    for(int i=0;i<10000;i++)arr[i]=arr[i%k];
                    w.println(10000);
                    for(int i=0;i<10000;i++)w.print(arr[i]+" ");
                    w.println();
                }
                
            }
            
            
        }
        w.close();
    }
}