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 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 __init__(self, file):
        self._file = file
        self._fd = file.fileno()
        self.writable = "x" in file.mode or "w" in file.mode
        self.write = super(FastIO, self).write if self.writable else None
 
    def _fill(self):
        s = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
        self.seek((self.tell(), self.seek(0,2), super(FastIO, self).write(s))[0])
        return s
 
    def read(self):
        while self._fill(): pass
        return super(FastIO,self).read()
 
    def readline(self):
        while self.newlines == 0:
            s = self._fill(); self.newlines = s.count(b"\n") + (not s)
        self.newlines -= 1
        return super(FastIO, self).readline()
 
    def flush(self):
        if self.writable:
            os.write(self._fd, self.getvalue())
            self.truncate(0), self.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')
 
if __name__ == '__main__':
    main()
#threading.Thread(target=main).start()