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:
import os,io
input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
cases = int(input())
for t in range(cases):
    n = int(input())
    b = list(map(int,input().split()))
    a = [0]*(2*n)
    d = {i:True for i in range(1,2*n+1)}

    for i in range(n):
        p = i+1
        pos = 2*p-1
        a[pos-1] = b[i]
        d[b[i]] = False

    f = 0
    for i in range(n):
        p = i+1
        pos = 2*p
        v = b[i]+1
        while v in d and not d[v]:
            v+=1
        if v>2*n:
            print(-1)
            f=1
            break
        else:
            d[v] = False
            a[pos-1] = v
    if f==0:
        print(*a)