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:
I = input
for _ in range(int(I())):
    try:
        n = int(I())
        b = list(map(int, I().split(' ')))
        a = [0 for i in range((2 * n) + 1)]
        s = []
        for i in range(n):
            if a[b[i]] == 0:
                a[b[i]] = 1
                s.append(str(b[i]))
                s.append('0')
            else:
                assert 1 == 0
        for i in range(n):
            t = 0
            for j in range(b[i] + 1, (2 * n) + 1):
                if a[j] == 0:
                    t = j
                    a[j] = 1
                    break
            if t == 0:
                assert 1 == 0
            s[2*(i+ 1)-1] = str(t)
        # print(s)
        for i in range(1, (2 * n) + 1):
            if a[i] == 0:
                assert 1 == 0
        for i in s:
            print(i,end=' ')
        print()
    except AssertionError:
        print('-1')