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:
# aadiupadhyay
import os.path
from math import gcd, floor, ceil
from collections import *
import sys
mod = 1000000007
INF = float('inf')
def st(): return list(sys.stdin.readline().strip())
def li(): return list(map(int, sys.stdin.readline().split()))
def mp(): return map(int, sys.stdin.readline().split())
def inp(): return int(sys.stdin.readline())
def pr(n): return sys.stdout.write(str(n)+"\n")
def prl(n): return sys.stdout.write(str(n)+" ")


if os.path.exists('input.txt'):
    sys.stdin = open('input.txt', 'r')
    sys.stdout = open('output.txt', 'w')


def solve():

    def update(ind):
        ind += n
        segment[ind] = -INF
        while ind > 1:
            ind >>= 1
            segment[ind] = max(segment[2*ind], segment[2*ind+1])

    def getMax(a, b):
        a += n
        b += n
        s = -INF
        while a <= b:
            if a % 2:
                s = max(s, segment[a])
                a += 1
            if b % 2 == 0:
                s = max(s, segment[b])
                b -= 1
            b >>= 1
            a >>= 1
        update(s-1)
        return s

    n = inp()
    l = li()
    ma = [0]
    segment = [0]
    ans1 = []
    for i in range(n):
        ma.append(0)
        segment.append(0)
        ans1.append(0)

    segment = segment + segment
    for i in range(n):
        segment[i+n] = i+1

    for i in range(n-1, 0, -1):
        segment[i] = max(segment[2*i], segment[2*i+1])
    ans2 = list(ans1)
    ma[l[0]] = 1
    ans1[0] = l[0]
    ans2[0] = l[0]
    i = 1
    update(l[0]-1)
    while i < n:
        if l[i] > l[i-1]:
            ma[l[i]] = 1
            update(l[i]-1)
            ans2[i] = l[i]
            ans1[i] = l[i]

        i += 1
    j = 1

    for i in range(n):
        if ans1[i] == 0:
            while j < n and ma[j] == 1:
                j += 1
            ans1[i] = j
            ma[j] = 1
    for i in range(n):
        if ans2[i] != 0:
            now = ans2[i]
        elif ans2[i] == 0:
            last = getMax(0, now-1)
            ans2[i] = last

    print(*ans1)
    print(*ans2)


for _ in range(inp()):
    solve()