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.File; 
import java.io.FileReader; 
import java.io.PrintWriter;
import java.io.FileOutputStream;
import java.io.BufferedOutputStream;
import java.io.FileNotFoundException;
public class First
{

	// ------------------------- MAIN FUNCITON ----------------------- // 
	public static void main( String[] args )				
	{

		// try 
		{

	        // Scanner sc = new Scanner(new FileReader("input.txt")); // INPUT FILE
	        // PrintWriter pw = new PrintWriter("output.txt");		  // OUTPUT FILE
	        Scanner sc = new Scanner( System.in ) ;			 // CONSOLE INPUT
	        PrintWriter pw = new PrintWriter( System.out ) ; // CONSOLE OUTPUT
	        int testcases = 0 ;
	        testcases = sc.nextInt( ) ;
	        while( (testcases --) != 0 )
	        {

	        	//CODE GOES HERE
	        	Integer n = sc.nextInt( ) ;
	        	ArrayList< Integer > v = new ArrayList<>( Collections.nCopies( n , 0 ) ) ;
	        	TreeSet< Integer > s = new TreeSet< >( ) ;
	        	TreeSet< Integer > ss = new TreeSet< >( ) ;
	        	for( int i = 0 ; i < n ; i++ )
	        	{
	        		s.add( i + 1 ) ;
	        		ss.add( i + 1 ) ;
	        	}
	        	for( int i = 0 ; i < n ; i++ )
	        	{
	        		v.set( i , sc.nextInt( ) ) ;
	        	}
	        	ArrayList< Integer > mx = new ArrayList<>( ) ;
	        	ArrayList< Integer > mn = new ArrayList<>( ) ;
	        	//calculating minimum 
	        	for( int i = 0 ; i < n ; i++ )
	        	{
	        		int val = 0 ;
	        		if( s.contains( v.get( i ) ) == true )
	        		{
	        			val = v.get( i ) ;
	        		}
	        		else
	        			val = s.first( ) ;
	        		mn.add( val ) ;
	        		s.remove( val ) ;
	        	}
	        	// printing minimum
	        	for( Integer i : mn )
	        	{
	        		pw.print( i + " " ) ;
	        	}
	        	pw.println( ) ;

	        	// calculation maximum
	        	for( int i = 0 ; i < n ; i++ )
	        	{
	        		int val = ss.floor( v.get( i ) ) ;
	        		mx.add( val ) ;
	        		ss.remove( val ) ;
	        	}
	        	// printing maximum
	        	for( Integer i : mx )
	        	{
	        		pw.print( i + " " ) ;
	        	}
	        	pw.println( ) ;

	        }
	        pw.close( ) ; 
	    } 
	    // catch (FileNotFoundException fnfe) 
	    // {
	    //     fnfe.printStackTrace();
	    // }
	}
}