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:
/*
*	Author: CrazyDoctor
*	
*	File: E.java
*	
*	Date: 13:43:47, 29.03.2021
*/

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

public class E {
	public static void main(String args[]) {
		FastScanner sc = new FastScanner();
		FastPrinter fp = new FastPrinter();
		
		int t = sc.nextInt();

		while(t-->0) {
			int n = sc.nextInt();
			int[] a = new int[n];
			int[] a1 = new int[n];
			boolean[] used = new boolean[n+1];
			boolean[] used1 = new boolean[n+1];
			TreeMap<Integer, Integer> mp = new TreeMap<>();
			for (int i = 1; i <= n; i++) mp.put(i, i);
			int prev = -1;
			for (int i = 0; i < n; i++) {
				int q = sc.nextInt();
				if (q != prev) {
					a[i] = q;
					a1[i] = q;
					prev = q;
					used[q] = true;
					used1[q] = true;
					mp.remove(q);
				}
			}
			int prevval = 1;
			
			for (int i = 1; i < n; i++) {
				if (a[i] == 0) {
					while (used[prevval]) prevval++;
					used[prevval] = true;
					a[i] = prevval;
				}
			}

			for (int i = 1; i < n; i++) {
				if (a1[i] == 0) {
					prev = a1[i - 1];
					int k = mp.lowerEntry(prev).getValue();
					a1[i] = k;
					mp.remove(k);
				}
			}
			
			fp.println(n, a);
			fp.println(n, a1);
		}
		

		fp.close();
	}

	static class FastPrinter extends PrintWriter {

		FastPrinter() {
			super(System.out);
		}

		void println(int n, int[] a) {
			for (int i = 0; i < n; i++) {
				this.print(a[i] + (i == n - 1 ? "\n" : " "));
			}
		}

		void println(int n, long[] a) {
			for (int i = 0; i < n; i++) {
				this.print(a[i] + (i == n - 1 ? "\n" : " "));
			}
		}

		void println(int n, double[] a) {
			for (int i = 0; i < n; i++) {
				this.print(a[i] + (i == n - 1 ? "\n" : " "));
			}
		}

		void println(ArrayList<Object> a) {
			int n = a.size();
			for (int i = 0; i < n; i++) {
				this.print(a.get(i) + (i == n - 1 ? "\n" : " "));
			}
		}
	}

	static class FastScanner {
		BufferedReader br;
		StringTokenizer st;
		
		public FastScanner() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}
		
		String next() {
			while (st == null || !st.hasMoreElements()) {
				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());
		}

		int[] nextIntA(int n) {
			int[] a = new int[n];
			for (int i = 0; i < n; i++)
				a[i] = this.nextInt();
			return a;
		}

		long[] nextLongA(int n) {
			long[] a = new long[n];
			for (int i = 0; i < n; i++)
				a[i] = this.nextLong();
			return a;
		}

		double[] nextDoubleA(int n) {
			double[] a = new double[n];
			for (int i = 0; i < n; i++)
				a[i] = this.nextDouble();
			return a;
		}

		ArrayList<Integer> nextIntAL(int n) {
			ArrayList<Integer> a = new ArrayList<Integer>(n);
			for (int i = 0; i < n; i++)
				a.add(this.nextInt());
			return a;
		}

		ArrayList<Long> nextLongAL(int n) {
			ArrayList<Long> a = new ArrayList<Long>(n);
			for (int i = 0; i < n; i++)
				a.add(this.nextLong());
			return a;
		}

		ArrayList<Double> nextDoubleAL(int n) {
			ArrayList<Double> a = new ArrayList<Double>(n);
			for (int i = 0; i < n; i++)
				a.add(this.nextDouble());
			return a;
		}

		ArrayList<String> nextStringAL(int n) {
			ArrayList<String> a = new ArrayList<String>(n);
			for (int i = 0; i < n; i++)
				a.add(this.next());
			return a;
		}
	}
}