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.*;
public class Main
{
	public static void main(String[] args)throws Exception
	{
	    BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
	    PrintWriter pw=new PrintWriter(System.out);
	    int t=Integer.parseInt(br.readLine());
	    while(t-->0)
	    {
	        int n=Integer.parseInt(br.readLine());
	        String s[]=(br.readLine()).split(" ");
	        int a[]=new int[n];
	        for(int i=0;i<n;i++)
	        {
	            a[i]=Integer.parseInt(s[i]);
	        }
	        int min[]=new int[n];
	        int max[]=new int[n];
	        min[0]=a[0];
	        max[0]=a[0];
	        boolean num[]=new boolean[n+1];
	        boolean num2[]=new boolean[n+1];
	        num[a[0]]=true;
	        num2[a[0]]=true;
	        
	        
	        
	        for(int i=1;i<n;i++)
	        {
	            if(a[i-1]!=a[i])
	            {
	                min[i]=a[i];
	                max[i]=a[i];
	                num[a[i]]=true;
	                num2[a[i]]=true;
	                
	            }
	        }
	        int pt=1;
	        for(int i=1;i<n;i++)
	        {
	            if(min[i]==0)
	            {
	                while(pt<=n && num[pt])
	                {
	                    pt++;
	                }
	                min[i]=pt;
	                num[pt]=true;
	            }
	        }
	        
	        LinkedList<Integer> ll=new LinkedList<Integer>();
	        int largest=0;
	        for(int i=0;i<n;i++)
	        {
	            if(max[i]!=0)
	            {
	                for(int j=largest+1;j<max[i];j++)
	                {
	                    ll.add(j);
	                }
	                largest=max[i];
	            }
	            else
	            {
	                max[i]=ll.pollLast();
	            }
	            
	        }
	        for(int i=0;i<n;i++)
	        {
	            pw.print(min[i]+" ");
	            
	        }
	        pw.println();
	        
	        for(int i=0;i<n;i++)
	        {
	            pw.print(max[i]+" ");
	            
	        }
	        pw.println();
	        
	    }
	    pw.flush();
	}
}