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 

def is_in_array(value, array):
    l = 0
    r = len(array) - 1
    while l < r:
        mid = (l + r) // 2
        if array[mid] == value:
            return True
        elif array[mid] > value:
            l = mid + 1
        else:
            r = mid - 1
    return False

t = int(input())
for _ in range(t):
    n = int(input())
    values = [int(i) for i in input().split()]
    sorted_values = sorted(values) # for binary search
    taken = [False] * (2 * n + 1)
    for x in values:
        taken[x] = True 
    ans = [None] * 2 * n
    for i in range(n):
        ans[2 * i] = values[i]
        for j in range(values[i] + 1, 2 * n + 1):
            if not taken[j]:
                ans[2 * i + 1] = j
                taken[j] = True
                break
        else:
            print(-1)
            break
    else:
        print(" ".join(map(str,ans)))