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.HashSet;

public class Main {

	public static void main(String[] args) {
		new Main();
	}

	public Main() {
		FastScanner fs = new FastScanner();
		java.io.PrintWriter out = new java.io.PrintWriter(System.out);
		solve(fs, out);
		out.flush();
		//if (fs.hasNext()) throw new AssertionError("read input");
	}

	public void solve(FastScanner fs, java.io.PrintWriter out) {
		int t = fs.nextInt();
		while(t --> 0) {
			int n = fs.nextInt(), k = fs.nextInt();
			HashSet<Integer> set = new HashSet<>();
			for (int i = 0;i < n;++ i) set.add(fs.nextInt());
			if (set.size() > k) {
				out.println(-1);
			} else {
				for (int i = 1;set.size() < k;++ i) set.add(i);
				out.println(n * k);
				Integer[] array = set.toArray(new Integer[k]);
				out.print(array[0]);
				for (int i = 1;i < n * k;++ i) out.print(" " + array[i % k]);
				out.println();
			}
		}
	}

	static class FastScanner {
		private final java.io.InputStream in = System.in;
		private final byte[] buffer = new byte[1024];
		private int ptr = 0;
		private int buflen = 0;
		private boolean hasNextByte() {
			if (ptr < buflen) return true;
			ptr = 0;
			try {
				buflen = in.read(buffer);
			} catch (java.io.IOException e) {
				e.printStackTrace();
			}
			return buflen > 0;
		}
		private byte readByte() {
			return hasNextByte() ? buffer[ptr++ ] : -1;
		}
		private static boolean isPrintableChar(byte c) {
			return 32 < c || c < 0;
		}
		private static boolean isNumber(int c) {
			return '0' <= c && c <= '9';
		}
		public boolean hasNext() {
			while (hasNextByte() && !isPrintableChar(buffer[ptr])) ptr++ ;
			return hasNextByte();
		}
		public String next() {
			if (!hasNext()) throw new java.util.NoSuchElementException();
			StringBuilder sb = new StringBuilder();
			byte b;
			while (isPrintableChar(b = readByte())) sb.appendCodePoint(b);
			return sb.toString();
		}
		public final long nextLong() {
			if (!hasNext()) throw new java.util.NoSuchElementException();
			long n = 0;
			try {
				byte b = readByte();
				if (b == '-') {
					while(isNumber(b = readByte())) n = n * 10 + '0' - b;
					return n;
				} else if (!isNumber(b)) throw new NumberFormatException();
				do n = n * 10 + b - '0'; while (isNumber(b = readByte()));
			} catch (java.util.NoSuchElementException e) {
			}
			return n;
		}
		public final int nextInt() {
			if (!hasNext()) throw new java.util.NoSuchElementException();
			int n = 0;
			try {
				byte b = readByte();
				if (b == '-') {
					while(isNumber(b = readByte())) n = n * 10 + '0' - b;
					return n;
				} else if (!isNumber(b)) throw new NumberFormatException();
				do n = n * 10 + b - '0'; while (isNumber(b = readByte()));
			} catch (java.util.NoSuchElementException e) {
			}
			return n;
		}
		public double nextDouble() {
			return Double.parseDouble(next());
		}
	}
	public static class Arrays {
		public static void sort(final int[] array) {
			int l, min = 0xFFFFFFFF, max = 0;
			for (l = 0;l < array.length;++ l) {
				int i = array[l];
				min &= i;
				max |= i;
				if ((i & 0x80000000) == 0) break;
			}
			for (int r = l + 1;r < array.length;++ r) {
				int i = array[r];
				min &= i;
				max |= i;
				if ((i & 0x80000000) != 0) {
					array[r] = array[l];
					array[l ++] = i;
				}
			}
			int use = min ^ max, bit = Integer.highestOneBit(use & 0x7FFFFFFF);
			if (bit == 0) return;
			sort(array, 0, l, use, bit);
			sort(array, l, array.length, use, bit);
		}
		private static void sort(final int[] array, final int left, final int right, final int use, int digit) {
			if (right - left <= 96) {
				for (int i = left + 1;i < right;++ i) {
					int tmp = array[i], tmp2, j;
					for (j = i;j > left && (tmp2 = array[j - 1]) > tmp;-- j) array[j] = tmp2;
					array[j] = tmp;
				}
				return;
			}
			int l = left;
			while(l < right && (array[l] & digit) == 0) ++ l;
			for (int r = l + 1;r < right;++ r) {
				int i = array[r];
				if ((i & digit) == 0) {
					array[r] = array[l];
					array[l ++] = i;
				}
			}
			if ((digit = Integer.highestOneBit(use & digit - 1)) == 0) return;
			sort(array, left, l, use, digit);
			sort(array, l, right, use, digit);
		}
		public static void sort(final long[] array) {
			int l;
			long min = 0xFFFFFFFFFFFFFFFFL, max = 0;
			for (l = 0;l < array.length;++ l) {
				long i = array[l];
				min &= i;
				max |= i;
				if ((i & 0x8000000000000000L) == 0) break;
			}
			for (int r = l + 1;r < array.length;++ r) {
				long i = array[r];
				min &= i;
				max |= i;
				if ((i & 0x8000000000000000L) != 0) {
					array[r] = array[l];
					array[l ++] = i;
				}
			}
			long use = min ^ max, bit = Long.highestOneBit(use & 0x7FFFFFFFFFFFFFFFL);
			if (bit == 0) return;
			sort(array, 0, l, use, bit);
			sort(array, l, array.length, use, bit);
		}
		private static void sort(final long[] array, final int left, final int right, final long use, long digit) {
			if (right - left <= 96) {
				for (int i = left + 1, j;i < right;++ i) {
					long tmp = array[i], tmp2;
					for (j = i;j > left && (tmp2 = array[j - 1]) > tmp;-- j) array[j] = tmp2;
					array[j] = tmp;
				}
				return;
			}
			int l = left;
			while(l < right && (array[l] & digit) == 0) ++ l;
			for (int r = l + 1;r < right;++ r) {
				long i = array[r];
				if ((i & digit) == 0) {
					array[r] = array[l];
					array[l ++] = i;
				}
			}
			if ((digit = Long.highestOneBit(use & digit - 1)) == 0) return;
			sort(array, left, l, use, digit);
			sort(array, l, right, use, digit);
		}
	}

	public static class IntMath {
		public static int gcd(int a, int b) {
			while (a != 0) if ((b %= a) != 0) a %= b; else return a;
			return b;
		}
		public static int gcd(int... array) {
			int ret = array[0];
			for (int i = 1;i < array.length;++ i) ret = gcd(ret, array[i]);
			return ret;
		}
		public static long gcd(long a, long b) {
			while (a != 0) if ((b %= a) != 0) a %= b; else return a;
			return b;
		}
		public static long gcd(long... array) {
			long ret = array[0];
			for (int i = 1;i < array.length;++ i) ret = gcd(ret, array[i]);
			return ret;
		}
		public static long lcm(long a, long b) {
			return a / gcd(a, b) * b;
		}
		public static int pow(int a, int b) {
			int ans = 1;
			for (int mul = a;b > 0;b >>= 1, mul *= mul) if ((b & 1) != 0) ans *= mul;
			return ans;
		}
		public static int pow(int a, int b, int mod) {
			if (b < 0) b = b % (mod - 1) + mod - 1;
			long ans = 1;
			for (long mul = a;b > 0;b >>= 1, mul = mul * mul % mod) if ((b & 1) != 0) ans = ans * mul % mod;
			return (int)ans;
		}
		public static int floorsqrt(long n) {
			return (int)Math.sqrt(n + 0.1);
		}
		public static int ceilsqrt(long n) {
			return n <= 1 ? (int)n : (int)Math.sqrt(n - 0.1) + 1;
		}
	}
}