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.io.*;
import java.math.*;
import java.util.*;
import java.util.stream.*;
import java.lang.management.*;
import static java.lang.Math.*;

@SuppressWarnings("unchecked")
public class P1348B {

  void add(Map<Integer, Integer> m, int a) {
    m.put(a, m.getOrDefault(a, 0) + 1);
  }

  void rem(Map<Integer, Integer> m, int a) {
    int c = m.get(a) - 1;

    if (c > 0) {
      m.put(a, c);
    } else {
      m.remove(a);
    }
  }

  public void run() throws Exception {
    for (int t = nextInt(); t > 0; t--) {
      int n = nextInt(), k = nextInt();
      TreeSet<Integer> as = new TreeSet();
      List<Integer> l = new ArrayList();
      for (int i = 0, a; i < n; a = nextInt(), l.add(a), as.add(a), i++);

      if (as.size() > k) {
        println(-1);
        continue;
      }

      Map<Integer, Integer> ac = new LinkedHashMap();
      for (int a : as) { add(ac, a); }
      for (int i = k - as.size(); i > 0; add(ac, as.first()), i--);

      println(k * n);
      for (int i = 0; i < n; i++) {
        for (Map.Entry<Integer, Integer> me : ac.entrySet()) {
          for (int j = me.getValue(); j > 0; print(me.getKey() + " "), j--);
        }
      }
      println();
    }
  }

  public static void main(String... args) throws Exception {
    br = new BufferedReader(new InputStreamReader(System.in));
    pw = new PrintWriter(new BufferedOutputStream(System.out));
    new P1348B().run();
    br.close();
    pw.close();
    System.err.println("\n[Time : " + (System.currentTimeMillis() - startTime) + " ms]");
    long gct = 0, gcc = 0;
    for (GarbageCollectorMXBean garbageCollectorMXBean : ManagementFactory.getGarbageCollectorMXBeans()) {
      gct += garbageCollectorMXBean.getCollectionTime();
      gcc += garbageCollectorMXBean.getCollectionCount();
    }
    System.err.println("[GC time : " + gct + " ms, count = " + gcc + "]");
    System.err.println("[JIT time : " + ManagementFactory.getCompilationMXBean().getTotalCompilationTime() + " ms]");
  }

  static long startTime = System.currentTimeMillis();
  static BufferedReader br;
  static PrintWriter pw;
  StringTokenizer stok;

  String nextToken() throws IOException {
    while (stok == null || !stok.hasMoreTokens()) {
      String s = br.readLine();
      if (s == null) { return null; }
      stok = new StringTokenizer(s);
    }
    return stok.nextToken();
  }

  void print(byte b) { print("" + b); }
  void print(int i) { print("" + i); }
  void print(long l) { print("" + l); }
  void print(double d) { print("" + d); }
  void print(char c) { print("" + c); }
  void print(Object o) {
    if (o instanceof int[]) { print(Arrays.toString((int [])o));
    } else if (o instanceof long[]) { print(Arrays.toString((long [])o));
    } else if (o instanceof char[]) { print(Arrays.toString((char [])o));
    } else if (o instanceof byte[]) { print(Arrays.toString((byte [])o));
    } else if (o instanceof short[]) { print(Arrays.toString((short [])o));
    } else if (o instanceof boolean[]) { print(Arrays.toString((boolean [])o));
    } else if (o instanceof float[]) { print(Arrays.toString((float [])o));
    } else if (o instanceof double[]) { print(Arrays.toString((double [])o));
    } else if (o instanceof Object[]) { print(Arrays.toString((Object [])o));
    } else { print("" + o); }
  }
  void printsp(int [] a) { for (int i = 0, n = a.length; i < n; print(a[i] + " "), i++); }
  void printsp(long [] a) { for (int i = 0, n = a.length; i < n; print(a[i] + " "), i++); }
  void print(String s) { pw.print(s); }
  void println() { println(""); }
  void println(byte b) { println("" + b); }
  void println(int i) { println("" + i); }
  void println(long l) { println("" + l); }
  void println(double d) { println("" + d); }
  void println(char c) { println("" + c); }
  void println(Object o) { print(o); println(); }
  void println(String s) { pw.println(s); }
  int nextInt() throws IOException { return Integer.parseInt(nextToken()); }
  long nextLong() throws IOException { return Long.parseLong(nextToken()); }
  double nextDouble() throws IOException { return Double.parseDouble(nextToken()); }
  char nextChar() throws IOException { return (char) (br.read()); }
  String next() throws IOException { return nextToken(); }
  String nextLine() throws IOException { return br.readLine(); }
  int [] readInt(int size) throws IOException {
    int [] array = new int [size];
    for (int i = 0; i < size; i++) { array[i] = nextInt(); }
    return array;
  }
  long [] readLong(int size) throws IOException {
    long [] array = new long [size];
    for (int i = 0; i < size; i++) { array[i] = nextLong(); }
    return array;
  }
  double [] readDouble(int size) throws IOException {
    double [] array = new double [size];
    for (int i = 0; i < size; i++) { array[i] = nextDouble(); }
    return array;
  }
  String [] readLines(int size) throws IOException {
    String [] array = new String [size];
    for (int i = 0; i < size; i++) { array[i] = nextLine(); }
    return array;
  }

  int gcd(int a, int b) {
    if (a == 0) return Math.abs(b); if (b == 0) return Math.abs(a);
    a = Math.abs(a); b = Math.abs(b);
    int az = Integer.numberOfTrailingZeros(a), bz = Integer.numberOfTrailingZeros(b);
    a >>>= az; b >>>= bz;
    while (a != b) {
      if (a > b) { a -= b; a >>>= Integer.numberOfTrailingZeros(a); }
            else { b -= a; b >>>= Integer.numberOfTrailingZeros(b); }
    }
    return (a << Math.min(az, bz));
  }

  long gcd(long a, long b) {
    if (a == 0) return Math.abs(b); if (b == 0) return Math.abs(a);
    a = Math.abs(a); b = Math.abs(b);
    int az = Long.numberOfTrailingZeros(a), bz = Long.numberOfTrailingZeros(b);
    a >>>= az; b >>>= bz;
    while (a != b) {
      if (a > b) { a -= b; a >>>= Long.numberOfTrailingZeros(a); }
            else { b -= a; b >>>= Long.numberOfTrailingZeros(b); }
    }
    return (a << Math.min(az, bz));
  }

  boolean isPrime(int v) {
    if (v <= 1) { return false; }
    if (v <= 3) { return true; }
    if (((v & 1) == 0) || ((v % 3) == 0)) { return false; }
    for (int m = 5, n = (int)sqrt(v); m <= n; m += 6) { if (((v % m) == 0) || ((v % (m + 2)) == 0)) { return false; } }
    return true;
  }

  int [] rshuffle(int [] a) { // RANDOM shuffle
    Random r = new Random();
    for (int i = a.length - 1, j, t; i >= 0; j = r.nextInt(a.length), t = a[i], a[i] = a[j], a[j] = t, i--);
    return a;
  }

  int [] qshuffle(int [] a) { // QUICK shuffle
    int m = new Random().nextInt(10) + 2;
    for (int i = 0, n = a.length, j = m % n, t; i < n; t = a[i], a[i] = a[j], a[j] = t, i++, j = (i * m) % n);
    return a;
  }

  long [] shuffle(long [] a) {
    Random r = new Random();
    for (int i = a.length - 1; i >= 0; i--) {
      int j = r.nextInt(a.length);
      long t = a[i]; a[i] = a[j]; a[j] = t;
    }
    return a;
  }

  void shuffle(Object [] a) {
    Random r = new Random();
    for (int i = a.length - 1; i >= 0; i--) {
      int j = r.nextInt(a.length);
      Object t = a[i]; a[i] = a[j]; a[j] = t;
    }
  }

  int [] sort(int [] a) {
    final int SHIFT = 16, MASK = (1 << SHIFT) - 1, SIZE = (1 << SHIFT) + 1;
    int n = a.length, ta [] = new int [n], ai [] = new int [SIZE];
    for (int i = 0; i < n; ai[(a[i] & MASK) + 1]++, i++);
    for (int i = 1; i < SIZE; ai[i] += ai[i - 1], i++);
    for (int i = 0; i < n; ta[ai[a[i] & MASK]++] = a[i], i++);
    int [] t = a; a = ta; ta = t;
    ai = new int [SIZE];
    for (int i = 0; i < n; ai[(a[i] >> SHIFT) + 1]++, i++);
    for (int i = 1; i < SIZE; ai[i] += ai[i - 1], i++);
    for (int i = 0; i < n; ta[ai[a[i] >> SHIFT]++] = a[i], i++);
    return ta;
  }

  void flush() {
    pw.flush();
  }

  void pause() {
    flush(); System.console().readLine();
  }
}