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:
# -*- coding: utf-8 -*-
"""
Created on Tue Sep 22 20:42:16 2020

@author: Dark Soul
"""

t=int(input(''))
for _ in range(t):
    n=int(input(''))
    tst=list(map(int,input().split()))
    b=[0]
    for i in tst:
        b.append(i)
    a=[0]*(2*n+1)
    used=[0]*(2*n+1)
    khaise=0
    for i in range(1,n+1):
        a[2*(i-1)+1]=b[i]
        if used[b[i]]:
            khaise=1
        used[b[i]]=1

    for i in range(2,2*n+1,2):
        for j in range(a[i-1],2*n+1,1):
            
            if (1-used[j]):
                break
        if j>2*n:
            khaise=1
        a[i]=j
        used[j]=1
    
    for i in range(1,2*n+1):
        if (1-used[i]):
            khaise=1
    if khaise:
        print(-1)
    else:
        print(*a[1:])