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 CPP):

Solution:
#pragma GCC optimize ("Ofast")
#include <bits/stdc++.h>
using namespace std;
#define lli long long int
#define FIO ios_base::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr);
#define FOR(j,n) for(int i=j;i<=n;i++)
#define REVERSE(j,n) for(int i=j;i>=n;i--)
#define fi first
#define se second
#define mod 1000000007
#define pb push_back
#define pii pair<int,int>
#define mp make_pair

lli A[10000000] = {0};	

void placeLeft(vector<int> &q,bool minimize)
{
	set<int> left;
	for(int i=1;i<=(int)q.size();i++)
		left.insert(i);
	for(int i:q)
	{
		if(i!=-1)
			left.erase(i);
	}
	int lastPlaced = -1;
	for(int &i : q)
	{
		if(i==-1)
		{
			set<int>::const_iterator it;
			if(minimize)
				it = left.begin();
			else
				it = --left.lower_bound(lastPlaced);
			i = *it;
			left.erase(it);
		}
		else
			lastPlaced = i;
	}
}

void solve()
{
	lli n;
	cin >> n;
	vector<int> q(n);
	FOR(0,n-1) cin >> q[i];
	vector<int> res1(n,-1), res2(n,-1);
	for(int i=0,lastPlaced = -1;i<n;lastPlaced = q[i],i++)
	{
		if(lastPlaced != q[i])
		{
			res1[i] = q[i];
			res2[i] = q[i];
		}
	}
	placeLeft(res1,true);
	placeLeft(res2,false);
	for(int x:res1)
		cout << x << " ";
	cout << "\n";
	for(int x:res2)
		cout << x << " ";
	cout << "\n";
	return;
}

int main()
{
	FIO;
	lli t;
	cin >> t;
	while(t--)
		solve();
	return 0;
}