You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.

Input

Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).

The first line of each test case consists of one integer n — the number of elements in the sequence b (1 ≤ n ≤ 100).

The second line of each test case consists of n different integers b_1, …, b_n — elements of the sequence b (1 ≤ b_i ≤ 2n).

It is guaranteed that the sum of n by all test cases doesn't exceed 100.

Output

For each test case, if there is no appropriate permutation, print one number -1.

Otherwise, print 2n integers a_1, …, a_{2n} — required lexicographically minimal permutation of numbers from 1 to 2n.

Example

Input


5
1
1
2
4 1
3
4 1 3
4
2 3 4 5
5
1 5 7 2 8


Output


1 2 
-1
4 5 1 2 3 6 
-1
1 3 5 6 7 9 2 4 8 10 
def min_number_after(number:int):
	global other_sequence
	for k in range(len(other_sequence)):
		if other_sequence[k] > number:
			return other_sequence[k]
	return "No"


def list_subtract():
	global sequence, n
	other_sequence = list(range(1, 2*n+1))
	for k in sequence:
		other_sequence.remove(k)
	return other_sequence


t = int(input())
for i in range(t):
	n = int(input())
	sequence = list(map(int, input().split(' ')))
	other_sequence = list_subtract()
	quantity_operations = len(sequence)*2
	for j in range(0, quantity_operations, 2):
		temp = min_number_after(sequence[j])
		if temp != "No":
			sequence.insert(j + 1, temp)
			other_sequence.remove(sequence[j + 1])
		else:
			print(-1)
			break
	else:
		sequence = [str(x) for x in sequence]
		print(' '.join(sequence))