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 
import heapq
def restorePermutation(n, A):
    ans = []
    unUsed = set(list(range(1, 2*n + 1)))
    for val in A:
        if not 1 <= val <= 2*n: return -1
        ans.append(val)
        ans.append(-1)
        unUsed.remove(val)
        
    minHeapq = [val for val in unUsed]
    heapq.heapify(minHeapq)
    #idea use BST in future
    #use heap for now
    for i in range(1, 2*n, 2):
        temp = []
        while minHeapq and minHeapq[0] < ans[i - 1]:
            temp.append(heapq.heappop(minHeapq))
        if not minHeapq or (minHeapq and minHeapq[0] < ans[i - 1]): 
            return -1
        ans[i] = heapq.heappop(minHeapq)
        while temp:
            heapq.heappush(minHeapq, temp.pop())
    
    return ans
        
        



testCases = int(input())
while testCases:
    n = int(input())
    A = [int(val) for val in input().split()]
    ans = restorePermutation(n, A)
    if ans == -1:
        print()
        print(ans)
    else:
        for val in ans:
            print(val, end = " ")
    testCases -= 1