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.io.PrintWriter;
import java.util.*;

public class RestoringThePermutation {
	public static void main(String[] args)  {
		FastScanner fs = new FastScanner();
		PrintWriter out = new PrintWriter(System.out);
		int testCase = fs.nextInt();
		for (int tc = 0; tc < testCase; tc++) {
			int n = fs.nextInt();
			int[] arr = fs.readArray(n);
			int[] minAns = new int[n];
			int[] maxAns = new int[n];
			Arrays.fill(minAns, -1);
			Arrays.fill(maxAns, -1);
			TreeSet<Integer> maxSet = new TreeSet<>();
			TreeSet<Integer> minSet = new TreeSet<>();
			for(int i = 1; i <= n; ++i) {
				maxSet.add(i);
				minSet.add(i);
			}
			minAns[0] = maxAns[0] = arr[0];
			maxSet.remove(arr[0]);
			minSet.remove(arr[0]);
			for(int i = 1; i < n; ++i) {
				if(arr[i - 1] != arr[i]) {
					maxAns[i] = minAns[i] = arr[i];
					maxSet.remove(arr[i]);
					minSet.remove(arr[i]);
				}
			}
			for(int i = 0; i < n; ++i) {
				if(minAns[i] == -1) {
					minAns[i] = minSet.first();
					minSet.remove(minAns[i]);
				}
				if(maxAns[i] == -1) {
					maxAns[i] = maxSet.floor(arr[i]);
					maxSet.remove(maxAns[i]);
				}
			}
			for(int i : minAns) {
				System.out.print(i + " ");
			}
			System.out.println();
			for(int i : maxAns ) {
				System.out.print(i + " ");
			}
			System.out.println();
		}
	}
	static void ruffleSort(int[] a) {
		int n = a.length;
		Random r = new Random();
		for(int i = 0; i < a.length; ++i) {
			int oi = r.nextInt(n), t = a[i];
			a[i] = a[oi];
			a[oi] = t;
		}
		Arrays.sort(a);
	}
	static void sort(int[] a)   {
		ArrayList<Integer> l = new ArrayList<>();
		for(int i : a) l.add(i);
		Collections.sort(l);
		for(int i = 0; i < a.length; ++i) a[i] = l.get(i);
	}
	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());
		}
		int[] readArray(int n) {
			int[] a = new int[n];
			for(int i = 0; i < n; ++i)
				a[i] = nextInt();
			return a;
		}
		long nextLong() {
			return Long.parseLong(next());
		}
	}
}