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.*;
import java.lang.*;
import java.util.*;

public class Task {
	public static void main(String[] args) {
		InputStream inputStream = System.in;
		OutputStream outputStream = System.out;
		InputReader in = new InputReader(inputStream);
		PrintWriter out = new PrintWriter(outputStream);

		Solution solution = new Solution();
		solution.solve(in, out);

		out.close();
	}

	static class InputReader {
		public BufferedReader reader;
		public StringTokenizer tokenizer;

		public InputReader(InputStream stream) {
			reader = new BufferedReader(new InputStreamReader(stream), 32768);
			tokenizer = null;
		}

		public String next() {
			while (tokenizer == null || !tokenizer.hasMoreTokens()) {
				try {
					tokenizer = new StringTokenizer(reader.readLine());
				} catch (IOException e) {
					throw new RuntimeException(e);
				}
			}
			return tokenizer.nextToken();
		}

		public int nextInt() {
			return Integer.parseInt(next());
		}

		public double nextDouble() {
			return Double.parseDouble(next());
		}

		public float nextFloat() {
			return Float.parseFloat(next());
		}

		public long nextLong() {
			return Long.parseLong(next());
		}

		public boolean nextBoolean() {
			return Boolean.parseBoolean(next());
		}

		public char nextChar() {
			return next().charAt(0);
		}

	}

	static class OrderedPair<T extends Comparable<T>, U extends Comparable<U>>
		        implements Comparable<OrderedPair<T, U>> {
		public T first;
		public U second;

		OrderedPair(T first, U second) {
			this.first = first;
			this.second = second;
		}

		@Override
		public int compareTo(OrderedPair<T, U> other) {
			if ((this.first).compareTo(other.first) != 0) {
				return (this.first).compareTo(other.first);
			}
			return (this.second).compareTo(other.second);
		}
	}



	static class Solution {
		public void solve(InputReader in, PrintWriter out) {
			int t = in.nextInt();
			while(t-- > 0) {
				int n = in.nextInt();
				int[] q = new int[n + 1];
				for(int i = 1; i <= n; ++i) q[i] = in.nextInt();
				LinkedList <Integer> nxt = new LinkedList<>();
				for(int i = 1; i <= n; ++i) {
					if(q[i] > q[i - 1]) {
						out.print(q[i]);
						out.print(" ");
						for(int j = q[i - 1] + 1; j < q[i]; ++j)  nxt.add(j);
					} else {
						out.print(nxt.peek());
						out.print(" ");
						nxt.pop();
					}
				}
				out.println();
				Stack <Integer> st=new Stack<>();
				for(int i = 1; i <= n; ++i) {
					if(q[i] > q[i - 1]) {
						out.print(q[i]);
						out.print(" ");
						for(int j = q[i - 1] + 1; j < q[i]; ++j)  st.add(j);
					} else {
						out.print(st.peek());
						out.print(" ");
						st.pop();
					}
				}
				out.println();
			}
		}
	}
}