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 sys import stdin
from collections import defaultdict
import bisect
input=stdin.readline
t=int(input())
for _ in range(t):
    n=int(input())
    a=list(map(int,input().split()))
    dict=defaultdict(int)
    arr=[]
    for i in range(n):
        arr.append([a[i],i])
        dict[a[i]]=1
    arr.sort()
    #print(arr)
    tmp=[]
    for i in range(1,2*n+1):
        if dict[i]==0:
            tmp.append(i)
    #print(tmp)
    fl=0
    farr={}
    for i in range(len(tmp)):
        if arr[i][0]>tmp[i]:
            fl=1
            break
    if fl==1:
        print(-1)
        continue
    # a, temp
    for x in a:
        val=tmp[bisect.bisect_right(tmp,x)]
        tmp.remove(val)
        print(x,val,end=" ")
    print()