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
# sys.stdin=open("input1.in","r")
# sys.stdout=open("outpul.out","w")
for _ in range(int(input())):
	N=int(input())
	L=list(map(int,input().split()))
	Hash=[0]*(2*N+1)
	for i in range(N):
		Hash[L[i]]+=1
	Nahi=[]
	for i in range(N):
		FLAG=0
		for j in range(L[i]+1,2*N+1):
			if Hash[j]==0:
				FLAG=1
				Nahi.append(L[i])
				Nahi.append(j)
				Hash[j]=1
				break
		if FLAG==0:
			break
	if FLAG==0:
		print("-1")
	else:
		for i in range(len(Nahi)):
			print(Nahi[i],end=" ")
		print()
	# Hai.sort()
	# Nahi.sort()
	# N=len(Nahi)
	# count=0
	# Pair=[0]*(2*N+1)
	# FLAG=0
	# print(Hai)
	# print(Nahi)
	# for i in Hai:
	# 	if count<N:
	# 		if Nahi[count]<i:
	# 			FLAG=1
	# 			break
	# 		Pair[i]=Nahi[count]
	# 		count+=1
	# 	else:
	# 		FLAG=1
	# 		break
	# if FLAG==1:
	# 	print("-1")
	# else:
	# 	for i in range(N):
	# 		print(L[i],end=" ")
	# 		print(Pair[L[i]],end=" ")
	# 	print()