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:
import bisect
import collections
import functools
import itertools
import math
import heapq
import random
import string


def repeat(_func=None, *, times=1):
    def decorator(func):
        @functools.wraps(func)
        def wrapper(*args, **kwargs):
            for _ in range(times):
                func(*args, **kwargs)

        return wrapper

    if _func is None:
        return decorator
    else:
        return decorator(_func)


def unpack(func=int):
    return map(func, input().split())


def l_unpack(func=int):
    """list unpack"""
    return list(map(func, input().split()))


def getint():
    return int(input())


def getmatrix(rows):
    return [list(map(int, input().split())) for _ in range(rows)]


def display_matrix(mat):
    for i in range(len(mat)):
        print(mat[i])


def getmin(arr, n):
    output = [-1] * n
    vstd = [False] * (n + 1)
    vstd[arr[0]] = True
    output[0] = arr[0]
    hp = [x for x in range(1, arr[0])]
    heapq.heapify(hp)
    maxi = arr[0] - 1
    heappush, heappop = heapq.heappush, heapq.heappop
    for i in range(1, n):
        if arr[i] > arr[i - 1]:
            output[i] = arr[i]
            vstd[arr[i]] = True
            while maxi + 1 < arr[i]:
                maxi += 1
                if not vstd[maxi]:
                    heappush(hp, maxi)
        else:
            while vstd[hp[0]]:
                heappop(hp)
            output[i] = heappop(hp)
            vstd[output[i]] = True
    return output


def getmax(arr, n):
    output = [-1] * n
    vstd = [False] * (n + 1)
    vstd[arr[0]] = True
    output[0] = arr[0]
    hp = [-x for x in range(1, arr[0])]
    heapq.heapify(hp)
    maxi = arr[0] - 1
    heappush, heappop = heapq.heappush, heapq.heappop
    for i in range(1, n):
        if arr[i] > arr[i - 1]:
            output[i] = arr[i]
            vstd[arr[i]] = True
            while maxi + 1 < arr[i]:
                maxi += 1
                if not vstd[maxi]:
                    heappush(hp, -maxi)
        else:
            while vstd[-hp[0]]:
                heappop(hp)
            output[i] = -heappop(hp)
            vstd[output[i]] = True
    return output


@repeat(times=int(input()))
def main():
    n = getint()
    arr = l_unpack()
    lmin = getmin(arr, n)
    print(*lmin)
    lmax = getmax(arr, n)
    print(*lmax)


MOD = 10 ** 9 + 7
main()