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 File {
	public static class FastScanner {
		BufferedReader br;
		StringTokenizer st;
		
		public FastScanner() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}
		
		String next() {
			while (st == null || !st.hasMoreElements()) {
				try {
					st = new StringTokenizer(br.readLine());
				}
				catch (IOException e) {
					e.printStackTrace();
				}
			}
			
			return st.nextToken();
		}
		
		int nextInt() {
			return Integer.parseInt(next());
		}
		
		long nextLong() {
			return Long.parseLong(next());
		}
	}
	public static void main(String[] args) {
		FastScanner sc = new FastScanner();
		PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
		int test=sc.nextInt();
		while(test>0)
		{
		    int n=sc.nextInt();
		    int arr[]=new int[n];
		    for(int i=0;i<n;i++)
		    {
		        arr[i]=sc.nextInt();
		    }
		    int max[]=new int[n];
		    int min[]=new int[n];
		    PriorityQueue<Integer> minheap=new PriorityQueue<>();
		    PriorityQueue<Integer> maxheap=new PriorityQueue<>(Collections.reverseOrder());
		    max[0]=arr[0];
		    min[0]=arr[0];
		    int p=0;
		    for(int i=1;i<arr[0];i++)
		    {
		        minheap.add(i);
		        maxheap.add(i);
		    }
		    for(int i=1;i<n;i++)
		    {
		        if(arr[i]!=arr[i-1])
		        {
		            max[i]=arr[i];
		            min[i]=arr[i];
		            for(int j=p+1;j<i;j++)
		            {
		                max[j]=maxheap.poll();
		                min[j]=minheap.poll();
		            }
		            for(int j=arr[i-1]+1;j<arr[i];j++){
		                minheap.add(j);
		                maxheap.add(j);
		            }
		            p=i;
		        }
		    }
		    for(int i=p+1;i<n;i++)
		    {
		        max[i]=maxheap.poll();
		        min[i]=minheap.poll();
		    }
		    for(int i=0;i<n;i++)out.print(min[i]+" ");
		    out.println();
		    for(int i=0;i<n;i++)out.print(max[i]+" ");
		    out.println();
		    test--;
		}
		out.close();
	}
}