human
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...

impo ... Can you provide a Python program to find the lexicographically minimal and lexicographically maximal permutations that could have been originally presented to Polycarp?
assistant
#Here is Python program to find the lexicographically minimal and lexicographically maximal permutations that could have been originally presented to Polycarp.
rt heapq

class Node:
    def __init__(self, key):
        self.left=None
        self.right=None
        self.data=key
    





















from math import *
from sys import *
from heapq import *
from collections import defaultdict
from itertools import permutations
import os, sys
from io import IOBase, BytesIO
#input =sys.stdin.buffer.readline
M=10**9+7


def main():
    
    t=int(input())
    for _ in range(t):
        n=int(input())
        q=list(map(int,input().split()))

        dup=list(set(q))
        dct=defaultdict(int)
        for x in dup:
            dct[x]=1
        arr=set()

        for i in range(1,n+1):
            if dct[i]==0:
                arr.add(i)
        arr=list(arr)
        heapq.heapify(arr)

        prev=-1
        # minperm=[]
        for x in q:
            if prev==-1:
                prev=x
                # minperm.append(x)
                print(x,end=" ")
            elif x==prev:
                val=heapq.heappop(arr)
                # minperm.append(val)
                print(val,end=" ")
            else:
                # minperm.append(x)
                print(x,end=" ")
                prev=x
        
        # print(*minperm)

        
        
        prev=-1
        print()
        # maxperm=[]

        maxq=[]
        heapq.heapify(maxq)

        for x in q:
            if prev==-1:
                prev=x
                # maxperm.append(x)
                print(x,end=" ")
                for jj in range(1,x):
                    heapq.heappush(maxq, -jj)
            elif x==prev:
                value=heapq.heappop(maxq)
                # maxperm.append(-value)
                print(-value,end=" ")
            else:
                #x>prev
                for jj in range(prev+1,x):
                    heapq.heappush(maxq, -jj)
                # maxperm.append(x)
                print(x,end=" ")
                prev=x
        
        # print(*maxperm)
        print()





BUFSIZE = 8192
class FastIO(BytesIO):
    newlines = 0
 
    def...