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.*;
import java.util.regex.*;
import java.util.stream.*;
public class Solution {
    public static void main(String[] args) throws Exception{
        BufferedReader bu = new BufferedReader(new InputStreamReader(System.in));
        int t = Integer.parseInt(bu.readLine());
        while (t > 0) {
            int n = Integer.parseInt(bu.readLine());
            String[] str = bu.readLine().split(" ");
            int[] q = new int[n];
            for (int i = 0; i < n; i++) {
                q[i] = Integer.valueOf(str[i]);
            }
            int[] p1 = new int[n], p2 = new int[n];
            PriorityQueue<Integer> inc = new PriorityQueue<>(), dec = new PriorityQueue<>((a,b) -> (b.compareTo(a)));
            for (int i = 0; i < n; i++) {
                if (i == 0 || q[i] > q[i-1]) {
                    p1[i] = q[i];
                    p2[i] = q[i];
                    if (i == 0) {
                        for (int j = q[i]-1; j > 0; j--) {
                            inc.add(j);
                            dec.add(j);
                        } 
                    } else {
                            for (int j = q[i]-1; j > q[i-1]; j--) {
                                inc.add(j);
                                dec.add(j);
                            }
                        }
                    }
                else {
                    p1[i] = inc.poll();
                    p2[i] = dec.poll();
                }
            }
            StringBuffer b = new StringBuffer(), c = new StringBuffer();
            for (int i = 0; i < n; i++) {
                if (i != n-1) {
                    b.append(p1[i] + " ");
                    c.append(p2[i] + " ");
                } else {
                    b.append(p1[i]);
                    c.append(p2[i]);
                }
            }
            System.out.printf("%s%n", b.toString());
            System.out.printf("%s%n", c.toString());
            --t;
        }
        bu.close();
    }
}