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.*;

public class Main
{

  
    public static void main(String args[])
    {
        FastScanner input = new FastScanner();
        StringBuilder ans = new StringBuilder();
        int tc = input.nextInt();
        while (tc-- > 0) {
         int n = input.nextInt();
        int a[] = new int[n];
        int small[] = new int[n];
        int big[] = new int[n];
        TreeSet<Integer> set = new TreeSet<>(Collections.reverseOrder());
        TreeSet<Integer> bset = new TreeSet<>(Collections.reverseOrder());
        for (int i = 0; i < n; i++) {
            a[i] = input.nextInt();
        }
        for (int i = 1; i <=n; i++) {
            set.add(i);
        }
       
        for (int i = 0; i < n; i++) {
            if(set.contains(a[i]))
            {
               small[i] = a[i];
               big[i] =a[i];
               set.remove(a[i]);
               
            }
        }
        bset = (TreeSet<Integer>) set.clone();
        
        
        for (int i = 0; i < n; i++) {
            if(small[i]==0)
            {
                small[i]=set.pollLast();
            }
        }
//        System.out.println(bset);
        for (int i = 0; i < n; i++) {
            if(big[i]==0)
            {
                for (Integer integer : bset) {
                    if(a[i]>=integer)
                    {
                        big[i] = integer;
                        bset.remove(integer);
                        break;
                    }
                }
            }
        }
        for (int i : small) {
            ans.append(i+" ");
        }
        ans.append("\n");
        for (int i : big) {
             ans.append(i+" ");
        }   
                ans.append("\n");

        }
        System.out.println(ans);
        
        
    }

    static class FastScanner
    {

        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer("");

        String next()
        {
            while (!st.hasMoreTokens()) {
                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() throws IOException
        {
            return br.readLine();
        }
    }

}