1315_C. Restoring Permutation
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 
```python
t = int(input())
from collections import defaultdict


for _ in range(t):
    dic = defaultdict(int)
    dic2 = defaultdict(int)
    n = int(input())
    bl = list(map(int,input().split()))
    for idx, b in enumerate(bl):
        dic[b] = idx + 1
    bl_sort = sorted(bl)
    s = 0
    kouho = []
    for b in range(1, 2*n+1):
        if not dic[b]:
            if len(kouho) == 0:
                print(-1)
                break
            min_idx = float("inf")
            pos = -1
            for k in kouho:
                if min_idx > dic[k]:
                    min_idx = dic[k]
                    pos = k
            dic2[pos] = b
            kouho.remove(pos)
        else:
            kouho.append(b)
            s += 1
    else:
        ansl = []
        for b in bl:
            ansl.append(b)
            ansl.append(dic2[b])
        print(*ansl)
    continue


```