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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.HashMap;
import java.util.Map;
import java.util.PriorityQueue;
import java.util.Set;
import java.util.Stack;
import java.util.StringTokenizer;
import java.util.TreeSet;

public class Main {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		FastReader s = new FastReader();
		int tc = s.nextInt();
		while(tc-- > 0) {
			int n = s.nextInt();
			boolean[] taken = new boolean[n + 1];
			int[] input = new int[n];
			for(int i = 0;i < n;i++) input[i] = s.nextInt();
			int min = 1;
			ArrayList<Integer> arr = new ArrayList<>();
			TreeSet<Integer> ts = new TreeSet<>();
			for(int i = 1;i <= n;i++) ts.add(i);
			arr.add(input[0]);
			taken[input[0]] = true;
			ts.remove(input[0]);
			for(int i = 1;i < n;i++) {
				if(input[i] != input[i - 1]) {
					taken[input[i]] = true;
					arr.add(input[i]);
					ts.remove(input[i]);
				}
				else {
					arr.add(ts.pollFirst());
				}
			}
			for(int i : arr) System.out.print(i + " ");
			System.out.println();
			ts = new TreeSet<>();
			for(int i = 1;i <= n;i++) ts.add(i);
			arr = new ArrayList<>();
			int max = input[0] - 1;
			taken = new boolean[n + 1];
			taken[input[0]] = true;
			arr.add(input[0]);
			ts = new TreeSet<>();
			for(int i = 0;i < n;i++) taken[input[i]] = true;
			for(int i = 1;i <= n;i++) {
				if(!taken[i]) ts.add(i);
			}
			for(int i = 1;i < n;i++) {
				if(input[i] != input[i - 1]) {
					taken[input[i]] = true;
					arr.add(input[i]);
				}
				else {
					int val = ts.floor(input[i]);
					ts.remove(val);
					arr.add(val);
				}
			}
			for(int i : arr) System.out.print(i + " ");
			System.out.println();
		}
	}
	// 3 2 1 4
	// 3 3 3 4

	// 3 2 4 1
	// 3 3 4 4
	// 3 1 4 2

	public static String[] findsub(String str , int si , int ei) {
		if(si > ei) {
			String[] arr = new String[0];
			return arr;
		}
		if(str.length() == 0) {
			String[] arr = new String[1];
			return arr;
		}
		// i have two options
		String[] first = findsub(str , si + 1 , ei);
		String[] second = findsub(str , si , ei - 1);
		String[] ans = new String[first.length + second.length + 1];
		ans[ans.length - 1] = str.substring(si, ei + 1);
		int i= 0;
		//System.out.println("fl = " + first.length + " sl = " + second.length);
		for(;i < first.length;i++) ans[i] = first[i]; // i = first.length
		for(;i < ans.length - 1;i++) ans[i] = second[i - first.length]; // first.length + second.length
		System.out.println("-----");
		for(String s : ans) System.out.println(s);

		return ans;

	}

	public static long fact(int n) {
		long mul = 1;
		for(int i = 1;i <= n;i++) mul *= i;
		return mul;
	}

	public static long nc2(long k) {
		return (k*(k - 1))/2;
	}

}
class store3 {
	long sumodd;
	long sumeven;
	int minodd = Integer.MAX_VALUE;
	int mineven = Integer.MAX_VALUE;
}
class FastReader 
{ 
	BufferedReader br; 
	StringTokenizer st; 

	public FastReader() 
	{ 
		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()); 
	} 

	double nextDouble() 
	{ 
		return Double.parseDouble(next()); 
	} 

	String nextLine() 
	{ 
		String str = ""; 
		try
		{ 
			str = br.readLine(); 
		} 
		catch (IOException e) 
		{ 
			e.printStackTrace(); 
		} 
		return str; 
	} 
}