Restore the Permutation
A sequence of $$$n$$$ numbers is called permutation if it contains all numbers from $$$1$$$ to $$$n$$$ exactly once. For example, the sequences [$$$3, 1, 4, 2$$$], [$$$1$$$] and [$$$2,1$$$] are permutations, but [$$$1,2,1$$$], [$$$0,1$$$] and [$$$1,3,4$$$] — are not.For a permutation $$$p$$$ of even length $$$n$$$ you can make an array $$$b$$$ of length $$$\frac{n}{2}$$$ such that:   $$$b_i = \max(p_{2i - 1}, p_{2i})$$$ for $$$1 \le i \le \frac{n}{2}$$$ For example, if $$$p$$$ = [$$$2, 4, 3, 1, 5, 6$$$], then:   $$$b_1 = \max(p_1, p_2) = \max(2, 4) = 4$$$  $$$b_2 = \max(p_3, p_4) = \max(3,1)=3$$$  $$$b_3 = \max(p_5, p_6) = \max(5,6) = 6$$$  As a result, we made $$$b$$$ = $$$[4, 3, 6]$$$.For a given array $$$b$$$, find the lexicographically minimal permutation $$$p$$$ such that you can make the given array $$$b$$$ from it.If $$$b$$$ = [$$$4,3,6$$$], then the lexicographically minimal permutation from which it can be made is $$$p$$$ = [$$$1,4,2,3,5,6$$$], since:   $$$b_1 = \max(p_1, p_2) = \max(1, 4) = 4$$$  $$$b_2 = \max(p_3, p_4) = \max(2, 3) = 3$$$  $$$b_3 = \max(p_5, p_6) = \max(5, 6) = 6$$$ A permutation $$$x_1, x_2, \dots, x_n$$$ is lexicographically smaller than a permutation $$$y_1, y_2 \dots, y_n$$$ if and only if there exists such $$$i$$$ ($$$1 \le i \le n$$$) that $$$x_1=y_1, x_2=y_2, \dots, x_{i-1}=y_{i-1}$$$ and $$$x_i<y_i$$$.
The first test case is parsed in the problem statement.
The first line of input data contains a single integer $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases.The description of the test cases follows.The first line of each test case contains one even integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10^5$$$).The second line of each test case contains exactly $$$\frac{n}{2}$$$ integers $$$b_i$$$ ($$$1 \le b_i \le n$$$) — elements of array $$$b$$$.It is guaranteed that the sum of $$$n$$$ values over all test cases does not exceed $$$2 \cdot 10^5$$$.
For each test case, print on a separate line:   lexicographically minimal permutation $$$p$$$ such that you can make an array $$$b$$$ from it;  or a number -1 if the permutation you are looking for does not exist. 
import sys
from bisect import bisect_right as bs
input = sys.stdin.readline
for _ in range(int(input())):
    n = int(input())
    b = list(map(int, input().split()))
    st = set(b)
    if len(st) != n // 2:
        print(-1)
        continue
    res = [i for i in range(1, n + 1) if i not in st]
    ans = []
    for i in range(n // 2 - 1, -1, -1):
        j = bs(res, b[-1]) - 1
        if j == -1:
            ans = -1
            break
        ans.extend((b[-1], res[j]))
        del b[-1]
        del res[j]
    if isinstance(ans, int):
        print(ans)
    else:
        sys.stdout.write(' '.join(map(str, ans[::-1])) + '\n')
from bisect import bisect_right as bs
for _ in range(int(input())):
    n = int(input())
    b = list(map(int, input().split()))
    st = set(b)
    if len(st) != n // 2:
        print(-1)
        continue
    res = [i for i in range(1, n + 1) if i not in st]
    ans = []
    for i in range(n // 2 - 1, -1, -1):
        j = bs(res, b[-1]) - 1
        if j == -1:
            ans = -1
            break
        ans.extend((b[-1], res[j]))
        del b[-1]
        del res[j]
    if isinstance(ans, int):
        print(ans)
    else:
        print(*ans[::-1])
from bisect import bisect_right as bs
for _ in range(int(input())):
    n = int(input())
    b = input().split()
    ans = [0] * n
    used = [0] * (n + 1)
    for (i, x) in enumerate(b):
        q = int(x)
        ans[2 * i + 1] = q
        used[q] = 1
    res = [i for i in range(1, n + 1) if not used[i]]
    if len(res) != n / 2:
        print(-1)
        continue
    for i in range(n // 2 - 1, -1, -1):
        j = bs(res, ans[2 * i + 1]) - 1
        if j == -1:
            ans = -1
            break
        ans[2 * i] = res[j]
        del res[j]
    if isinstance(ans, int):
        print(ans)
    else:
        print(*ans)