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 PYTHON):

Solution:
from collections import Counter, defaultdict, deque
import bisect
from sys import stdin, stdout
from itertools import repeat
import math


def inp(force_list=False):
    re = map(int, raw_input().split())
    if len(re) == 1 and not force_list:
        return re[0]
    return re

def inst():
    return raw_input().strip()

def gcd(x, y):
   while(y):
       x, y = y, x % y
   return x


mod = int(1e9)+7
ans = 0
def quickm(a, b):
    base = a
    re = 1
    while b:
        if b&1:
            re *= base
            re %= mod
        b >>= 1
        base *= base
        base %= mod
    return re

def inv(num):
    return quickm(num, mod-2)


def my_main():
    kase = inp()
    pans = []
    for _ in range(kase):
        n = inp()
        da = inp(1)
        mi = [da[0]]
        use = [0] * n
        for i in range(n):
            use[da[i]-1] = 1
        p = 0
        for i in range(1, n):
            if da[i]>da[i-1]:
                mi.append(da[i])
            else:
                while use[p]:
                    p += 1
                use[p] = 1
                mi.append(p+1)
        pans.append(' '.join(map(str, mi)))
        mx = [da[0]]
        use = [0] * n
        for i in range(n):
            use[da[i]-1] = 1
        p = da[0] - 2
        qq = []
        tq = []
        while p>=0 and not use[p]:
            use[p] = 1
            tq.append(p)
            p -= 1
        for j in reversed(tq):
            qq.append(j)
        for i in range(1, n):
            if da[i]>da[i-1]:
                mx.append(da[i])
                p = da[i] - 2
                tq = []
                while p>=0 and not use[p]:
                    use[p] = 1
                    tq.append(p)
                    p -= 1
                for j in reversed(tq):
                    qq.append(j)
            else:
                mx.append(qq[-1]+1)
                qq.pop()
        # print 111111111        tq = []
        while p>=0 and not use[p]:
            use[p] = 1
            tq.append(p)
            p -= 1
        for j in reversed(tq):
            qq.append(j)
        pans.append(' '.join(map(str, mx)))







    print '\n'.join(pans)
# print 4 **7
my_main()