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.Arrays;
import java.util.StringTokenizer;

/**
 * @Author:
 * @Date: 2021/4/24 23:31
 */
public class RestoringPermutations {
    static int [] nums = new int[2*100000+3];
    static int [] stk =  new int[2*100000+3];
    static boolean [] vis = new boolean[2*100000+3];
    public static void main(String[] args) {
        FastScanner fs = new FastScanner();
        int t = fs.nextInt();
        while ((t--)>0){
            int n = fs.nextInt();
            for(int i = 1; i <= n; i++){
                nums[i] = fs.nextInt();
                vis[i] = false;
            }
            int idx = 1;
            for(int i = 1; i <= n; i ++){
                if(nums[i]>nums[i-1]){
                    System.out.print(nums[i]+" ");
                    vis[nums[i]] = true;
                }else{
                    while (vis[idx]){
                        idx++;
                    }
                    System.out.print(idx+" ");
                    idx++;
                }
            }
            System.out.println();
            int k = 1;
            for(int i = 1; i <= n; i++){
                if(nums[i]>nums[i-1]){
                    System.out.print(nums[i]+" ");
                    for(int j = nums[i-1]+1; j < nums[i]; j++){
                        stk[k++] = j;
                    }
                }else{
                    System.out.print(stk[--k]+" ");
                }
            }
            System.out.println();
        }
    }
    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());
        }
    }
}