Problem: 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 

Solution (in PYTHON3):

Solution:
import sys
import math
import itertools
import functools
import collections
import operator
import fileinput
import copy
from collections import *
 
ORDA = 97  # a
def ii(): return int(input())
def mi(): return map(int, input().split())
def li(): return [int(i) for i in input().split()]
def lcm(a, b): return abs(a * b) // math.gcd(a, b)
def revn(n): return str(n)[::-1]
def dd(): return collections.defaultdict(int)
def ddl(): return collections.defaultdict(list)
def sieve(n):
    if n < 2: return list()
    prime = [True for _ in range(n + 1)]
    p = 3
    while p * p <= n:
        if prime[p]:
            for i in range(p * 2, n + 1, p):
                prime[i] = False
        p += 2
    r = [2]
    for p in range(3, n + 1, 2):
        if prime[p]:
            r.append(p)
    return r
def divs(n, start=2):
    r = []
    for i in range(start, int(math.sqrt(n) + 1)):
        if (n % i == 0):
            if (n / i == i):
                r.append(i)
            else:
                r.extend([i, n // i])
    return r
def divn(n, primes):
    divs_number = 1
    for i in primes:
        if n == 1:
            return divs_number
        t = 1
        while n % i == 0:
            t += 1
            n //= i
        divs_number *= t
def prime(n):
    if n == 2: return True
    if n % 2 == 0 or n <= 1: return False
    sqr = int(math.sqrt(n)) + 1
    for d in range(3, sqr, 2):
        if n % d == 0: return False
    return True
def convn(number, base):
    new_number = 0
    while number > 0:
        new_number += number % base
        number //= base
    return new_number
def cdiv(n, k): return n // k + (n % k != 0)
def ispal(s): # Palindrome
    for i in range(len(s) // 2 + 1):
        if s[i] != s[-i - 1]:
            return False
    return True
   
# a = [1,2,3,4,5] -----> print(*a) ----> list print krne ka new way

def function(arr, size,dp):
	if size == 1 or size == 0:
		return 1
	key = ''.join(map(str, arr[:size]))
	if dp[key]:
		return dp[key]
	output = function(arr, size-1, dp)
	key = ''.join(map(str, arr[:size-1]))
	dp[key] = output
	if (arr[size-2]*10 + arr[size-1]) <= 26:
		output += function(arr, size-2, dp)
		key = ''.join(map(str, arr[:size-2]))
		dp[key] = output
	return output

	
def main():
	for _ in range(ii()):
		n = ii()
		arr = li()
		b =[0]*2*n
		d = [True]*(2*n + 1)
		i = 0
		for ele in arr:
			d[ele] = False
			b[i] = ele
			i += 2
		ok = True
		for i in range(1,2*n,2):
			e = b[i-1] + 1
			while e <= 2*n:
				if d[e]:
					d[e] = False
					b[i] = e
					break
				e += 1
			else:
				ok = False
				break
		if ok:
			print(*b)
		else:
			print(-1)
main()