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.*;
import java.io.*;

public class Restoring_The_Permutation {

	static class FastReader {
		BufferedReader br;
		StringTokenizer st;

		public FastReader() {
			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());
		}

		String nextLine() {
			String str = "";
			try {
				str = br.readLine();
			} catch (IOException e) {
				e.printStackTrace();
			}
			return str;
		}

	}

	public static void shuffle(int[] a) {
		Random r = new Random();

		for (int i = 0; i <= a.length - 2; i++) {
			int j = i + r.nextInt(a.length - i);

			swap(a, i, j);
		}

		Arrays.sort(a);
	}

	public static void swap(int[] a, int i, int j) {
		int temp = a[i];
		a[i] = a[j];
		a[j] = temp;
	}

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		FastReader t = new FastReader();
		PrintWriter o = new PrintWriter(System.out);
		int test = t.nextInt();

		while (test-- > 0) {
			int n = t.nextInt();
			int[] a = new int[n];
			int[] small = new int[n];
			int[] big = new int[n];
			List<int[]> step = new ArrayList<>();
			int cur = 0;
			boolean[] s1 = new boolean[n + 1];
			boolean[] s2 = new boolean[n + 1];
			PriorityQueue<Integer> p1 = new PriorityQueue<>((x, y) -> x - y);
			PriorityQueue<Integer> p2 = new PriorityQueue<>((x, y) -> y - x);

			step.add(new int[] { -1, 0 });

			for (int i = 0; i < n; ++i)
				a[i] = t.nextInt();

			for (int i = 0; i < n; ++i)
				if (a[i] > cur) {
					small[i] = big[i] = cur = a[i];
					s1[cur] = s2[cur] = true;

					step.add(new int[] { i, cur });
				}

			step.add(new int[] { n, n + 1 });

			for (int i = 2; i < step.size(); ++i) {
				int[] low = step.get(i - 2);
				int[] high = step.get(i - 1);
				int[] then = step.get(i);
				++high[0];
				int h1 = high[1] - 1;

				while (h1 > low[1]) {
					if (!s1[h1]) {
						p1.add(h1);

						s1[h1] = true;
					}

					if (!s2[h1]) {
						p2.add(h1);

						s2[h1] = true;
					}

					--h1;
				}

				while (high[0] < then[0]) {
					int v1 = p1.remove();
					int v2 = p2.remove();
					small[high[0]] = v1;
					big[high[0]] = v2;

					++high[0];
				}
			}

			for (int v : small)
				o.print(v + " ");

			o.println();

			for (int v : big)
				o.print(v + " ");

			o.println();
		}

		o.flush();
		o.close();
	}
}