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.*;
import java.util.*;

public class Main {

    public static void main(String[] args) {
        BufferedReader bf = new BufferedReader( new InputStreamReader(System.in));
        int t = 0;
        try {
            t = Integer.parseInt(bf.readLine());
        } catch (IOException e) {
            e.printStackTrace();
        }
        for (int iTest = 0; iTest < t; iTest++) {
            int n = 0;
            try {
                n = Integer.parseInt(bf.readLine());
            } catch (IOException e) {
                e.printStackTrace();
            }
            List<Integer> perm = new ArrayList<>(n);
            List<String> tokens = null;
            try {
                tokens = Arrays.asList(bf.readLine().split(" "));
            } catch (IOException e) {
                e.printStackTrace();
            }
            for (int i = 0; i < n; i++) {
                perm.add(Integer.parseInt(tokens.get(i)));
            }
            TreeSet<Integer> active = new TreeSet<>();
            for (int i = 1; i <= n; i++) {
                active.add(i);
            }
            List<Integer> minAnswer = new ArrayList<>();
            int pos = 0;
            do {
                int st = pos;
                while (pos + 1 < n && perm.get(pos).equals(perm.get(pos + 1))) {
                    pos++;
                }
                active.remove(perm.get(pos));
                minAnswer.add(perm.get(pos));
                for (int j = st + 1; j <= pos; j++) {
                    minAnswer.add(active.pollFirst());
                }
                pos++;
            } while (pos < n);

            for (int i = 1; i <= n; i++) {
                active.add(i);
            }
            List<Integer> maxAnswer = new ArrayList<>(n);
            for (int i = 0; i < n; i++) {
                maxAnswer.add(0);
            }
            pos = 0;
            do {
                int st = pos;
                while (pos + 1 < n && perm.get(pos).equals(perm.get(pos+1))) {
                    pos++;
                }
                active.remove(perm.get(pos));
                maxAnswer.set(st, perm.get(pos));
                for (int j = st + 1; j <= pos; j++) {
                    Integer lower = active.lower(perm.get(pos));
                    maxAnswer.set(j, active.lower(perm.get(pos)));
                    active.remove(lower);
                }
                pos++;
            } while (pos < n);

            for (Integer value : minAnswer) {
                System.out.print(value + " ");
            }
            System.out.println();
            for (Integer value : maxAnswer) {
                System.out.print(value + " ");
            }
            System.out.println();
        }
    }
}