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:
from bisect import bisect_left as bl, bisect_right as br, insort
import sys
import heapq
from math import *
from collections import defaultdict as dd, deque
def data(): return sys.stdin.readline().strip()
def mdata(): return map(int, data().split())
#sys.setrecursionlimit(100000)

for i in range(int(input())):
    n=int(data())
    b=list(mdata())
    l=[]
    for i in range(1,2*n+1):
        if i not in b:
            l.append(i)
    l1=[]
    for i in range(n):
        flag = False
        for j in range(len(l)):
            if b[i]<l[j]:
                l1.append(b[i])
                l1.append(l[j])
                l.pop(j)
                flag=True
                break
        if flag==False:
            break
    if flag==True:
        print(*l1)
    else:
        print(-1)