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:
def solution():
    for _ in range(int(input())):
        n = int(input())
        arr = [int(i) for i in input().split()]
        maximal = max(arr)
        minimal = min(arr)
        mandatory = set(range(1, minimal+maximal+1))
        if len(mandatory) > 2*n or minimal > 1:
            print(-1)
            continue
        if len(mandatory) != 2*n:
            mandatory.update(set(range(len(mandatory)+1, (2*n)+1)))
        mandatory.difference_update(set(arr))
        ans = [0]*n
        for i in range(0,n,1):
            if arr[i]+1 in mandatory:
                ans[i] = arr[i]+1
                mandatory.remove(ans[i])
            else:
                for j in mandatory:
                    if j > arr[i]:
                        ans[i] = j
                        mandatory.remove(j)
                        break
        if 0 in ans:
            print(-1)
            continue
        from functools import reduce
        from operator import add
        print(*reduce(add, zip(arr, ans)))
solution()