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.*;

public class Solution
{
    public static void main(String []args)
    {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
       
        while(t-- > 0)
        {
          int n = sc.nextInt();
          int arr[] = new int[n];
          for(int i = 0 ; i < n ; i++)
          {
              arr[i] = sc.nextInt();
          }
          int pos[] = new int[n];
          int pp = 0;
          while(pp < n)
          {
              int j = pp;
              while(j+1 < n && arr[j]==arr[j+1])
              {
                  j++;
              }
              pos[pp] = 1;
              
              pp = j+1;
          }
          TreeMap<Integer,Integer> map = new TreeMap<Integer,Integer>();
          StringBuffer str = new StringBuffer("");
          
          for(int i = 1 ; i <= n ; i++)
          {
              map.put(i,1);
          }
          for(int i = 0 ; i < n ; i++)
          {
              if(pos[i] == 1)
              {
                  str.append(arr[i] + " ");
                  map.remove(arr[i]);
              }
              
              else
              {
                  int x = map.firstKey();
                  str.append(x + " ");
                  map.remove(x);
              }
          }
          System.out.println(str);
          
          for(int i = 1 ; i <= n ; i++)
          {
              map.put(i,1);
          }
          
          int post[] = new int[n];
          for(int i = n-1 ; i >= 0 ; i--)
          {
              if(pos[i] == 1)
              post[i] = i;
              
              else
              {
                  if(i==n-1)
                  post[i] = -1;
                  
                  else
                  post[i] = post[i+1];
              }
          }
          StringBuffer str2 = new StringBuffer("");
          int gr = -1;
          for(int i = 0 ; i < n ; i++)
          {
              if(pos[i] == 1)
              {
                  str2.append(arr[i] + " ");
                  map.remove(arr[i]);
              }
              
              else
              {
                  int x = 0;
                  if(post[i] == -1)
                  x = map.lastKey();
                  
                  else
                  x = map.lowerKey(arr[i]);
                  
                  
                 
                  
                  str2.append(x + " ");
                  
                  map.remove(x);
              }
          }
          System.out.println(str2);
        }
    }
}