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 javax.print.attribute.standard.MediaSize.Other;


/**
 *
 * @author Saju
 *
 */

public class Main {

	private static int dx[] = { 1, 0, -1, 0 };
	private static int dy[] = { 0, -1, 0, 1 };

	private static final long INF = Long.MAX_VALUE;
	private static final int INT_INF = Integer.MAX_VALUE;
	private static final long NEG_INF = Long.MIN_VALUE;
	private static final int NEG_INT_INF = Integer.MIN_VALUE;
	private static final double EPSILON = 1e-10;

	private static final int MAX = 107;
	
	private static final long MOD = 1000000007;

	private static final int MAXN = 10000007;
	private static final int MAXA = 10000009;
	private static final int MAXLOG = 22;

	public static void main(String[] args) throws IOException {

		InputReader in = new InputReader(System.in);
		PrintWriter out = new PrintWriter(System.out);

//		 InputReader in = new InputReader(new FileInputStream("src/test.in"));
//		 PrintWriter out = new PrintWriter(new BufferedWriter(new FileWriter("src/test.out")));



/*

1
5 3
1 2 2 1 3



1
10 9
1 1 1 1 1 1 1 1 1 1

1 
2 1
1 1

*/
		
		int test = in.nextInt();
		for (int t = 1; t <= test; t++) {
			int n = in.nextInt();
			int k = in.nextInt();
			int arr[] = new int[n];
			
			Set<Integer> set = new HashSet<>();
			int max = 0;
			int[] count = new int[MAX];
			int min = INT_INF;
			int minV = arr[0];
			int maxV = arr[0];
			for(int i = 0; i < n; i++) {
				arr[i] = in.nextInt();
				set.add(arr[i]);
				count[arr[i]]++;
				if(max < count[arr[i]]) {
					max = count[arr[i]];
					maxV = arr[i];
				}
				max = max(max, count[arr[i]]);
				if(min > count[arr[i]]) {
					min = count[arr[i]];
					minV = arr[i];
				}
			}
			if(set.size() > k) {
				out.println("-1");
				continue;
			}
			if(k == n) {
				out.println(n);
				for(int i = 0; i < n; i++) {
					out.print(arr[i] + " ");
				}
				out.println();
				continue;
			}
//			if(set.size() == k) {
//				List<Integer> list = new ArrayList<>();
//				for(int i = 0; i < max; i++) {
//					for(Integer e : set) {
//						list.add(e);
//					}
//				}
//				int sz = list.size();
//				out.println(sz);
//				for(int i = 0; i < sz; i++) {
//					out.print(list.get(i) + " ");
//				}
//				out.println();
//				continue;
//			}
			
			int sz = set.size();
//			System.out.println(minV);
			List<Integer> s = new ArrayList<Integer>(set);
			for(int i = sz + 1; i <= k; i++) {
				if(minV != maxV) {
					count[minV]++;
				}
				s.add(minV);
				max = max(max, count[minV]);
			}
//			System.out.println(max);
			max = 101;
//			System.out.println(max);
			List<Integer> list = new ArrayList<>();
			for(int i = 1; i <= max; i++) {
				for(Integer e : s) {
					list.add(e);
				}
			}
			sz = list.size();
			out.println(sz);
			for(int i = 0; i < sz; i++) {
				out.print(list.get(i) + " ");
			}
			out.println();
			
			
			
		}
		
		
		
		out.flush();
		out.close();
		System.exit(0);
	}
	
	private static long bigMod(long n, long k, long m) {

		long ans = 1;
		while (k > 0) {
			if ((k & 1) == 1) {
				ans = (ans * n) % m;
			}
			n = (n * n) % m;
			k >>= 1;
		}
		return ans;
	}
	
	private static int ceil(int n, int x) {
		int div = n / x;
		if(div * x != n) {
			div++;
		}
		return div;
	}

	private static class Point {
		double x;
		double y;

		Point(double x, double y) {
			this.x = x;
			this.y = y;
		}

		double getDistance(Point p2) {
			Point p1 = this;
			double dx = p1.x - p2.x;
			double dy = p1.y - p2.y;

			return Math.sqrt((dx * dx) + (dy * dy));
		}
	}

	private static int abs(int x) {
		if (x < 0) {
			return -x;
		}
		return x;
	}

	private static long abs(long x) {
		if(x < 0) {
			return -x;
		}
		return x;
	}

	private static int lcm(int a, int b) {
		return (a * b) / gcd(a, b);
	}

	private static int gcd(int a, int b) {
		if (a == 0)
			return b;
		return gcd(b % a, a);
	}

	private static int log(long x, int base) {
		return (int) (Math.log(x) / Math.log(base));
	}

	private static long min(long a, long b) {
		if (a < b) {
			return a;
		}
		return b;
	}

	private static int min(int a, int b) {
		if (a < b) {
			return a;
		}
		return b;
	}

	private static long max(long a, long b) {
		if (a < b) {
			return b;
		}
		return a;
	}

	private static int max(int a, int b) {
		if (a < b) {
			return b;
		}
		return a;
	}

	private static class InputReader {
		public BufferedReader reader;
		public StringTokenizer tokenizer;

		public InputReader(InputStream stream) {
			reader = new BufferedReader(new InputStreamReader(stream));
			tokenizer = null;
		}

		public String next() {
			try {
				while (tokenizer == null || !tokenizer.hasMoreTokens()) {
					tokenizer = new StringTokenizer(reader.readLine());

				}
			} catch (IOException e) {
				return null;
			}
			return tokenizer.nextToken();
		}

		public String nextLine() {
			String line = null;
			try {
				tokenizer = null;
				line = reader.readLine();
			} catch (IOException e) {
				throw new RuntimeException(e);
			}
			return line;
		}

		public int nextInt() {
			return Integer.parseInt(next());
		}

		public double nextDouble() {
			return Double.parseDouble(next());
		}

		public long nextLong() {
			return Long.parseLong(next());
		}

		public boolean hasNext() {
			try {
				while (tokenizer == null || !tokenizer.hasMoreTokens()) {
					tokenizer = new StringTokenizer(reader.readLine());
				}
			} catch (Exception e) {
				return false;
			}
			return true;
		}
	}
}