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:
/*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
*/
import java.util.*;
import java.io.*;

public class Main {
    static void runFromFile() {
        final String IODir = "D:/Kien/competitive_programming/io";
        final File inputFile = new File(IODir + "/input.txt");
        final File outputFile = new File(IODir + "/output.txt");
        try {
            System.setIn(new FileInputStream(inputFile));
            System.setOut(new PrintStream(outputFile));
        } catch (FileNotFoundException e) {
            e.printStackTrace();
        }
    }
    public static void main(String[] args) {      
        //runFromFile();
        Scanner inp = new Scanner(System.in);
        //PrintWriter out = new PrintWriter(System.out);
        int numTests = inp.nextInt();
        while (numTests-- > 0) {
        	int n = inp.nextInt();
        	int[] arr = new int[n];
        	for (int i = 0; i < n; i++) arr[i] = inp.nextInt();
        		
        	TreeMap<Integer, Boolean> set = new TreeMap<>();
        	for (int i = 1; i <= n; i++) set.put(i, true);
        	for (int i = 0; i < n; i++) set.remove(arr[i]);
        	int[] ans = new int[n];
        	for (int i = 0; i < n; i++) {
        		if (i == 0 || arr[i] != arr[i - 1]) {
        			ans[i] = arr[i];
        		}
        		else {
        			ans[i] = set.firstKey();
        			set.remove(ans[i]);
        		}
        	}
        	for (int i = 0; i < n; i++) System.out.print(ans[i] + (i == n - 1? "\n": " "));

        	set.clear();
        	for (int i = 1; i <= n; i++) set.put(i, true);
        	for (int i = 0; i < n; i++) set.remove(arr[i]);
        	ans = new int[n];
        	for (int i = 0; i < n; i++) {
        		if (i == 0 || arr[i] != arr[i - 1]) {
        			ans[i] = arr[i];
        		}
        		else {
        			Integer cur = set.lowerKey(arr[i]);
        			if (cur == null) cur = set.lastKey();
        			ans[i] = cur;
        			set.remove(ans[i]);
        		}
        	}

        	for (int i = 0; i < n; i++) System.out.print(ans[i] + (i == n - 1? "\n": " "));
        }
        
        //out.close();
    }
}