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.Collections;
import java.util.PriorityQueue;
import java.util.StringTokenizer;

public class Round710Div3E {
	public static void main(String[] args) throws NumberFormatException, IOException {
		BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
		int numT = Integer.parseInt(f.readLine());
		for (int i = 0; i < numT; i++) {
			int n = Integer.parseInt(f.readLine());
			StringTokenizer st = new StringTokenizer(f.readLine());
			int[] arr = new int[n];
			int[] minResult = new int[n];
			int[] maxResult = new int[n];
			int[] divs = new int[n+1];
			int count=0;
			PriorityQueue<Integer> mins = new PriorityQueue<>();
			PriorityQueue<Integer> maxs = new PriorityQueue<>(Collections.reverseOrder());
			boolean[] mark = new boolean[n+1];
			for (int j = 0; j < n; j++) {
				arr[j]=Integer.parseInt(st.nextToken());
				if(j==0 || arr[j]!=arr[j-1]) {
					minResult[j]=arr[j];
					maxResult[j]=arr[j];
					divs[count]=j;
					mark[arr[j]]=true;
					count++;
				}
			}
			divs[count]=n;
			for (int j = 1; j <= n; j++) {
				if(!mark[j])
					mins.offer(j);
			}
			
			for (int j = 0; j < n; j++) {
				if(j==0 || arr[j]!=arr[j-1]) 
					continue;
				minResult[j]=mins.poll();
			}
			
			for (int j = 0; j < count; j++) {
				int p = 1;
				if(j-1>=0)
					p=arr[divs[j-1]]+1;
				for (; p < arr[divs[j]]; p++) {
					maxs.offer(p);
				}
				for (int k = divs[j]+1; k < divs[j+1]; k++) {
					maxResult[k]=maxs.poll();
				}
			}
			
			
			
			for (int j = 0; j < n - 1; j++) {
				System.out.print(minResult[j] + " ");
			}
			System.out.println(minResult[n - 1]);
			
			for (int j = 0; j < n - 1; j++) {
				System.out.print(maxResult[j] + " ");
			}
			System.out.println(maxResult[n - 1]);
		}
	}
}