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 
#Pye
from os import path
from sys import stdin, stdout
from bisect import bisect_left, bisect_right
maxn = 200005
dd = [0 for i in range(maxn)]
a = [0 for i in range(maxn)]
if path.exists('inp.txt'):
    stdin = open("inp.txt", "r")
q = int(stdin.readline())
for _ in range(q):
    s = []; check = 0;
    n = int(stdin.readline())
    res = [0 for i in range(2*n+1)]
    for i in range(1, 2*n+1): dd[i] = 0
    a[1:] = list(map(int, stdin.readline().split()))
    for i in a: dd[i] += 1
    for i in range(1, 2*n+1):
        if not dd[i]: s.append(i)
    for i in range(1, n+1):
        if bisect_left(s, a[i]) != len(s):
            res[2*i-1] = a[i]
            res[2*i] = s[bisect_left(s, a[i])]
            s.pop(bisect_left(s, a[i]))
        else: check = 1; break
    if not check: print(*res[1:])
    else: print(-1)

