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:
import sys
import bisect
input = sys.stdin.readline
ins = lambda: input().rstrip()
ini = lambda: int(input().rstrip())
inm = lambda: map(int, input().rstrip().split())
inl = lambda: list(map(int, input().split()))
out = lambda x: print('\n'.join(map(str, x)))

finalans = []
t = ini()
for _ in range(t):
    n = ini()
    b = inl()
    a = [i for i in range(1, n*2+1) if i not in b]
    out = True
    j = 0
    for index, number in enumerate(sorted(b)):
        if number > a[index]:
            out = False
            finalans.append([-1])
            break
    if not out:
        continue
    ans = []
    for i in b:
        x = bisect.bisect_right(a, i)
        ans.append(i)
        ans.append(a[x])
        a.pop(x)
    finalans.append(ans)
for i in finalans:
    print(*i)