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 math
from sys import stdin as c


n=int(input())
a=[]
for i in range(n):
	input()
	a.append(list(map(int,c.readline().split(' '))))
def higher(ar,val):
	u=len(ar)-1
	if u==0:
		return 0
	if u==-1:
		return -1
	elif ar[u//2]>val:
		return higher(ar[:u//2+1],val)
	else:
		
		u1=higher(ar[u//2+1:],val)
		if u1>-1:
			u1+=u//2+1
		return u1
for i in a:
	n=len(i)
	s=set(i)
	n2=set(range(1,2*n+1))
	n2-=s
	d=dict(zip(i,range(n)))
	n2=list(n2)
	if n!=len(n2):
		print(-1)
		continue
	odp=[]
	b=True
	for j in range(n):
		h=higher(n2,i[j])
		if i[j]<n2[h]:
			odp.append(i[j])
			odp.append(n2[h])
			del n2[h]
		else:
			print(-1)
			b=False
			break
	if b:
		print(*odp)