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 JAVA):

Solution:


	import java.util.*;
	import java.io.*;
	import java.text.DecimalFormat;


	public class restore {


	static DecimalFormat df1 = new DecimalFormat("#.000000");
	static DecimalFormat df = new DecimalFormat("#.######");
	static boolean printing=false;

	static int N;

	static long total=0L;

	public static void main(String[] args) throws Exception {
		  
			
		 BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
			
		  // Scanner in = new Scanner(System.in);
		   
		   if(printing){
			   //in = new Scanner(new File("p3.in"));
			   in = new BufferedReader(new FileReader("restore.in"));
		   }
		   StringTokenizer tokenizer = new StringTokenizer(in.readLine());
			int cases = Integer.parseInt(tokenizer.nextToken());
			//tokenizer = new StringTokenizer(in.readLine());
			
			for(int j=1;j<=cases;j++){	
			
				tokenizer = new StringTokenizer(in.readLine());
			
				int n = Integer.parseInt(tokenizer.nextToken());
				HashMap<Integer, Integer> freq=new HashMap<>();
				ArrayList<Integer> F=new ArrayList<>();
				if(j==3578){
					//System.out.println("AAAAA"+(in.readLine()).replaceAll("\\s","+")+"BBB"+n);
				
					//continue;
				}
				int[] A   = new int[n];
				int[] Amax= new int[n];
				int[] Amin= new int[n];
				tokenizer = new StringTokenizer(in.readLine());
				
				TreeSet<Integer> availMax=new TreeSet<>();
				TreeSet<Integer> availMin=new TreeSet<>();

				for(int i=1;i<n;i++){
					availMax.add(i);
					availMin.add(i);
				}
				
				for(int i=0;i<n;i++){
					A[i] = Integer.parseInt(tokenizer.nextToken());
					if(i==0){
						Amax[0]=A[0];
						Amin[0]=A[0];
						availMax.remove(A[0]);
						availMin.remove(A[0]);
					}else{
						if(A[i]==A[i-1]){
							int temp1=availMax.floor(A[i]);
							Amax[i]=temp1;
							availMax.remove(temp1);
							
							/////////////////
							Amin[i]=availMin.pollFirst();
														
							
						} else {
							Amax[i]=A[i];
							Amin[i]=A[i];
							availMax.remove(A[i]);
							availMin.remove(A[i]);							
						}
					}

				}
				
				StringBuilder sMin=new StringBuilder();
				StringBuilder sMax=new StringBuilder();
				
				for(int i=0;i<n;i++){
					if(i!=n-1){
						sMin.append(Amin[i]).append(' ');
						sMax.append(Amax[i]).append(' ');
					}else{
						sMin.append(Amin[i]);
						sMax.append(Amax[i]);
					}

				}
				System.out.println(sMin);
				System.out.println(sMax);
			}
			in.close();
			
		}



	}