Problem: A permutation is a sequence of n integers from 1 to n, in which all numbers occur exactly once. For example, [1], [3, 5, 2, 1, 4], [1, 3, 2] are permutations, and [2, 3, 2], [4, 3, 1], [0] are not.

Polycarp was presented with a permutation p of numbers from 1 to n. However, when Polycarp came home, he noticed that in his pocket, the permutation p had turned into an array q according to the following rule: 

  * q_i = max(p_1, p_2, …, p_i). 



Now Polycarp wondered what lexicographically minimal and lexicographically maximal permutations could be presented to him.

An array a of length n is lexicographically smaller than an array b of length n if there is an index i (1 ≤ i ≤ n) such that the first i-1 elements of arrays a and b are the same, and the i-th element of the array a is less than the i-th element of the array b. For example, the array a=[1, 3, 2, 3] is lexicographically smaller than the array b=[1, 3, 4, 2].

For example, if n=7 and p=[3, 2, 4, 1, 7, 5, 6], then q=[3, 3, 4, 4, 7, 7, 7] and the following permutations could have been as p initially: 

  * [3, 1, 4, 2, 7, 5, 6] (lexicographically minimal permutation); 
  * [3, 1, 4, 2, 7, 6, 5]; 
  * [3, 2, 4, 1, 7, 5, 6]; 
  * [3, 2, 4, 1, 7, 6, 5] (lexicographically maximum permutation). 



For a given array q, find the lexicographically minimal and lexicographically maximal permutations that could have been originally presented to Polycarp.

Input

The first line contains one integer t (1 ≤ t ≤ 10^4). Then t test cases follow.

The first line of each test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5).

The second line of each test case contains n integers q_1, q_2, …, q_n (1 ≤ q_i ≤ n).

It is guaranteed that the array q was obtained by applying the rule from the statement to some permutation p.

It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5.

Output

For each test case, output two lines: 

  * on the first line output n integers — lexicographically minimal permutation that could have been originally presented to Polycarp; 
  * on the second line print n integers — lexicographically maximal permutation that could have been originally presented to Polycarp; 

Example

Input


4
7
3 3 4 4 7 7 7
4
1 2 3 4
7
3 4 5 5 5 7 7
1
1


Output


3 1 4 2 7 5 6 
3 2 4 1 7 6 5 
1 2 3 4 
1 2 3 4 
3 4 5 1 2 7 6 
3 4 5 2 1 7 6 
1 
1 

Solution (in PYTHON3):

Solution:
# DEFINING SOME GOOD STUFF
import sys
from math import *
import threading
from heapq import *
from itertools import count
from pprint import pprint
from collections import defaultdict

'''
    intialise defaultdict by any kind of value by default you want to take ( int -> 0 | list -> [] )
'''
from heapq import heapify, heappop, heappush

sys.setrecursionlimit(300000)
# threading.stack_size(10**8)
'''
-> if you are increasing recursionlimit then remember submitting using python3 rather pypy3
-> sometimes increasing stack size don't work locally but it will work on CF
'''

mod = 10 ** 9+7
inf = 10 ** 15
decision = ['NO', 'YES']
yes = 'YES'
no = 'NO'

# ------------------------------FASTIO----------------------------
import os

from io import BytesIO, IOBase

BUFSIZE = 8192


class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n")+(not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)


class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")


sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")


# _______________________________________________________________#

def npr(n, r):
    return (factorial(n)%mod) // (factorial(n-r)%mod) if n >= r else 0


def ncr(n, r):
    return (factorial(n)%mod) // ((factorial(r)%mod) * (factorial(n-r)%mod)) if n >= r else 0


def lower_bound(li, num):
    answer = -1
    start = 0
    end = len(li)-1

    while (start <= end):
        middle = (end+start) // 2
        if li[middle] >= num:
            answer = middle
            end = middle-1
        else:
            start = middle+1
    return answer  # min index where x is not less than num


def upper_bound(li, num):
    answer = -1
    start = 0
    end = len(li)-1

    while (start <= end):
        middle = (end+start) // 2

        if li[middle] <= num:
            answer = middle
            start = middle+1

        else:
            end = middle-1
    return answer  # max index where x is not greater than num


def abs(x):
    return x if x >= 0 else -x


def binary_search(li, val):
    # print(lb, ub, li)
    ans = -1
    lb = 0
    ub = len(li)-1
    while (lb <= ub):
        mid = (lb+ub) // 2
        # print('mid is',mid, li[mid])
        if li[mid] > val:
            ub = mid-1
        elif val > li[mid]:
            lb = mid+1
        else:
            ans = mid  # return index
            break
    return ans


def kadane(x):  # maximum sum contiguous subarray
    sum_so_far = 0
    current_sum = 0
    for i in x:
        current_sum += i
        if current_sum < 0:
            current_sum = 0
        else:
            sum_so_far = max(sum_so_far, current_sum)
    return sum_so_far


def pref(li):
    pref_sum = [0]
    for i in li:
        pref_sum.append(pref_sum[-1]+i)
    return pref_sum


def SieveOfEratosthenes(n):
    prime = [{1, i} for i in range(n+1)]
    p = 2
    while (p <= n):
        for i in range(p * 2, n+1, p):
            prime[i].add(p)
        p += 1
    return prime


def primefactors(n):
    factors = []
    while (n % 2 == 0):
        factors.append(2)
        n //= 2
    for i in range(3, int(sqrt(n))+1, 2):  # only odd factors left
        while n % i == 0:
            factors.append(i)
            n //= i
    if n > 2:  # incase of prime
        factors.append(n)
    return factors


def prod(li):
    ans = 1
    for i in li:
        ans *= i
    return ans


def sumk(a, b):
    print('called for', a, b)
    ans = a * (a+1) // 2
    ans -= b * (b+1) // 2
    return ans

def sumi(n):
    ans = 0
    if len(n) > 1:
        for x in n:
            ans += int(x)
        return ans
    else:
        return int(n)

# _______________________________________________________________#


# def main():
karmanya = int(input())
# karmanya = 1
# divisors = SieveOfEratosthenes(200010)
# print(divisors)
while karmanya != 0:
    karmanya -= 1
    n = int(input())
    # n,m,x= map(int, input().split())
    # n,t= map(int, input().split())
    # k = int(input())
    # a = [int(x) for x in input()]
    a = list(map(int, input().split()))
    # b = list(map(int, input().split()))
    # c = list(map(int, input().split()))
    # d = defaultdict(int)
    # s = list(input())
    maxheap = []
    minheap = []
    ans, ans1 = [a[0]], [a[0]]
    for i in range(1, a[0]):
        heappush(minheap, i)
        heappush(maxheap, -i)
    heapify(maxheap)
    heapify(minheap)
    for i in range(1, n):
        if a[i] == a[i-1]:
            ans.append(heappop(minheap))
            ans1.append(-heappop(maxheap))
        else:
            for j in range(a[i-1]+1, a[i]):
                heappush(minheap, j)
                heappush(maxheap, -j)
            ans.append(a[i])
            ans1.append(a[i])
    print(*ans)
    print(*ans1)



















































































































# t = threading.Thread(target=main)
# t.start()
# t.join()