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:
for _ in range(int(input())):
    n = int(input())
    b = [int(i) for i in input().split()]
    missed = []
    sorted_b = sorted(b)
    for ind, num in enumerate(sorted_b):
        if ind != 0:
            missed.extend(list(range(sorted_b[ind - 1] + 1, num)))
        else:
            missed.extend(list(range(1, num)))
    if sorted_b[-1] != 2 * n:
        missed.extend((range(sorted_b[-1] + 1, 2 * n + 1)))
    # print(missed)
    ans = []
    flag = 0
    for num in b:
        m = float('inf')
        for candidate in missed:
            if num < candidate < m:
                m = candidate
        if m != float('inf'):
            del missed[missed.index(m)]
        else:
            flag = 1
            break
        if min(m, num) == num:
            ans.extend([num, m])
    if flag == 0:
        print(*ans)
    else:
        print(-1)