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 sys

input = lambda: sys.stdin.readline().rstrip()


def last_proverka(v):
    for i in range(len(v)):
        i += 1
        if i not in v:
            return -1
    return True


# v - ans
# h  - b


def l_proverka(v, c):
    for i in range(len(c)):
        if v[i] == c[i]:
            return False
    return True


def x(m, sr):
    for i in range(1000):
        if i > sr and i not in m and i not in a:
            return i
    return False


for _ in range(int(input())):
    n = int(input())
    b = list(map(int, input().split()))
    a = [0] * 2 * n

    for i in range(n):
        a[2 * i] = b[i]
    k = True
    for i in range(2 * n):
        if a[i] == 0:
            c = x(b, a[i - 1])
            if c == False:
                k = False
                break
            else:
                a[i] = c
    if last_proverka(a) == True:
        print(*a)
    else:
        print(-1)
