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 PYTHON):

Solution:

from math import factorial as fac
from collections import defaultdict
from copy import deepcopy
import sys, math
f = None
try:
	f = open('q1.input', 'r')
except IOError:
	f = sys.stdin
if 'xrange' in dir(__builtins__):
	range = xrange
# print(f.readline())
# sys.setrecursionlimit(10**6)


def print_case_iterable(case_num, iterable):
	print("Case #{}: {}".format(case_num," ".join(map(str,iterable))))

def print_case_number(case_num, iterable):
	print("Case #{}: {}".format(case_num,iterable))

def print_iterable(A):
	print (' '.join(A))

def read_int():
	return int(f.readline().strip())
def read_int_array():
	return [int(x) for x in f.readline().strip().split(" ")]
def rns():
	a =  [x for x in f.readline().split(" ")]
	return int(a[0]), a[1].strip()
def read_string():
	return list(f.readline().strip())

NUMBER = 10**9 + 7
# NUMBER = 998244353
def factorial(n) : 
	M = NUMBER
	f = 1
  
	for i in range(1, n + 1):  
		f = (f * i) % M # Now f never can  
						# exceed 10^9+7  
	return f
def mult(a,b):
	return (a * b) % NUMBER

def minus(a , b):
	return (a - b) % NUMBER

def plus(a , b):
	return (a + b) % NUMBER

def egcd(a, b):
	if a == 0:
		return (b, 0, 1)
	else:
		g, y, x = egcd(b % a, a)
		return (g, x - (b // a) * y, y)

def modinv(a):
	m = NUMBER
	g, x, y = egcd(a, m)
	if g != 1:
		raise Exception('modular inverse does not exist')
	else:
		return x % m
from collections import deque, defaultdict 
import heapq

def bfs(V, E):
	s = set()
	print('\n'.join(map(str,E.items())))
	cnt = 0 
	for v in V:
		if v in s:
			continue
		q = deque([])
		s.add(v)
		q.append(v)
		while len(q) > 0:
			x = q.popleft()
			for e in E[x]:
				if e not in s:
					s.add(e)
					q.append(e)
			print(q)
		cnt += 1
	return cnt
		
# from typing import List, Tuple

def binary_count_smaller(L, x):
	l = 0
	if x < L[0]:
		return 0
	r = len(L)
	while r - l > 1:
		mid = (l+r)//2
		if L[mid] < x:
			l = mid
		else:
			r = mid
	return l + 1
def binary_count_larger(L, x):
	l = 0
	if x > L[-1]:
		return 0
	r = len(L)
	while r - l > 1:
		mid = (l+r)//2
		if L[mid] <= x:
			l = mid
		else:
			r = mid
	return len(L) - (r-1)

def regular_bs(L,x,l=None,r=None):
	l = 0 if l == None else l
	r = len(L) if r == None else r
	while (r-l) > 1:
		m = (r+l)//2
		l = m if L[m] <= x else l
		r = m if L[m] > x else r
	cnt = 0
	return l

from itertools import product

def preprocess():
	nums = dict(zip(sequences,range(10)))
	changes = defaultdict(dict)
	for comb in map(lambda x: ''.join(x), product(*([['0','1']]*7))):
		for seq in sequences:
			a = int(comb,2)
			b = int(seq,2)
			num_changes = bin(a^b).count('1')
			if num_changes == seq.count('1') - comb.count('1'):
				changes[comb][num_changes] = nums[seq]
	return changes


# import StringIO use for PYPY when lots of prints
#example usage:
#	 for l in res:
#	 	print >>output, str(len(l)) + ' ' +  ' '.join(l)
from collections import deque

# def solution1(x,y,l,r):
# 	pass
# 	if x*y*r == 0:
# 		return 0
# 	x = x % 2**r.bit_length()
# 	y = y % 2**r.bit_length()
# 	if l <= (x | y) <=r:
# 		return x | y
# 	binr = bi(r)
# 	binl = bi(l) #remeber l might be 0
# 	first_diff = r.bit_length() - (r^l).bit_length()
# 	res = binr[:first_diff]
# 	two_first = 2**(r^l).bit_length()
# 	use_zero = (x|y) & (two_first-1)
# 	if x&two_first and y&two_first:
# 		return res + solution(x,y,int(binr[first_diff+1:],2))
# 	if x&two_first:
# 		return max(use_zero, short(y,))
# 	# use_one = 
# 	# max((x|y)&(2**(r^l).bit_length()-1))


	# return x | y
def short(x,r):
	x = x % (2**(r.bit_length()))
	return x

def max_short(x,r):
	x = short(x,r)
	if x <= r:
		return x
	le = r.bit_length()-1
	return 2**le + max_short(x ,r%(2**le))

def bi(x):
	return bin(x)[2:]




def operation(x,y,z):
	return (x&z) * (y&z)

def get_min_z(x,y,z):
	return (x|y) & z

def solution(A,n,k):
	s = set(A)
	list_a = sorted(list(s))
	if len(s) > k:
		print(-1)
		return 
	to_repeat = []
	for i in range(k):
		to_repeat.append(list_a[i%len(list_a)])
	s_to_repeat = ' '.join(map(str,to_repeat))
	m = n * k
	final_list = [s_to_repeat for i in range(n)]
	final_s = ' '.join(final_list)
	print(m)
	print(final_s)

	


def main():
	T = read_int()
	for i in range(T):
		n,k = read_int_array()
		A = read_int_array()
		solution(A,n,k)


if __name__ == '__main__':
	main()