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 
from typing import List
import collections
import bisect
import itertools
import functools
from fractions import gcd
import heapq
from math import ceil, sqrt, floor

# import sys
# sys.setrecursionlimit(50000)


def bisect_ceil(A, target):
    low = 0
    high = len(A) - 1
    if not A or A[-1] < target:
        return -1
    if A[0] > target:
        return 0
    while low <= high:
        mid = (low + high) // 2
        midv = A[mid]
        if midv == target:
            return mid
        elif target > midv:
            low = mid + 1
        elif mid - 1 >= 0 and A[mid-1] < target < A[mid]:
            return mid
        else:
            high = mid - 1


t = int(input())
for _ in range(t):
    n = int(input())
    B = list(map(int, input().split()))
    B_set = set(B)
    candidate = [i for i in range(1, 2*n+1) if i not in B_set]

    result = [0] * (2 * n)
    p = ""
    for i in range(n):
        result[2*i] = B[i]
        index = bisect_ceil(candidate, B[i])
        if index == -1:
            p = '-1'
            break
        result[2*i+1] = candidate[index]
        del candidate[index]
    if p:
        print(p)
    else:
        print(' '.join([str(i) for i in result]))
