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 
#! /usr/bin/env python
# -*- coding: utf-8 -*-
# vim:fenc=tf-8
#

"""
"""

from operator import itemgetter
from collections import Counter

def solve(n, b):
    numset = set(range(1,2*n))
    sb = set(b)
    if len(sb) != n or not(numset >= sb):
        print(-1)
    else:
        used = [0]*(2*n+1)
        for bb in b:
            used[bb] = 1
        ans = []
        for bb in b:
            flag = True
            ans.append(bb)
            for i in range(bb+1,2*n+1):
                if used[i] == 0:
                    used[i] = 1
                    ans.append(i)
                    flag = False
                    break
            if flag:
                print(-1)
                return 0
        print(" ".join(map(str,ans)))
            

    


def main():
    t= int(input())
    for i in range(t):
        n = int(input())
        b = list(map(int,input().split()))
        solve(n, b)


if __name__ == "__main__":
    main()