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.util.*;

public class cf1506_Div2_E {
	
	
	public static void main(String args[]) throws IOException {
		FastScanner in = new FastScanner(System.in);
		PrintWriter out = new PrintWriter(System.out);
		
		int t = in.nextInt();
		for ( ; t > 0; t--) {
			int n = in.nextInt();
	
			int[] min = new int[n];
			int[] max = new int[n];
			
			
			int[] vals = new int[n];
			for (int i = 0; i < n; i++)
				vals[i] = in.nextInt();
			
			min[0] = vals[0];
			max[0] = vals[0];
			
			boolean[] fixed = new boolean[n + 1];
			fixed[vals[0]] = true;
			
			for (int i = 1; i < n; i++) {
				if (vals[i] != vals[i - 1]) {
					min[i] = vals[i];
					max[i] = vals[i];
					fixed[vals[i]] = true;
				}
			}
			
			
			TreeSet<Integer> minSet = new TreeSet<Integer>();
			TreeSet<Integer> maxSet = new TreeSet<Integer>();
			
			for (int i = 1; i <= n; i++) {
				if (fixed[i]) continue;
				minSet.add(i);
				maxSet.add(i);
			}
			
			for (int i = 1; i < n; i++) {
				if (vals[i] == vals[i - 1]) {
					int small = minSet.pollFirst();
					int large = maxSet.lower(vals[i]);
					maxSet.remove(large);
					min[i] = small;
					max[i] = large;
				}
			}
			
			for (int i = 0; i < n; i++) {
				out.print(min[i] + " ");
			}
			out.println();
			for (int i = 0; i < n; i++) {
				out.print(max[i] + " ");
			}
			out.println();
		}
		
		out.close();
	}
	
	static class FastScanner {
	    BufferedReader br;
	    StringTokenizer st;
		
	    public FastScanner(InputStream i) {
	        br = new BufferedReader(new InputStreamReader(i));
	        st = new StringTokenizer("");
	    }
				
	    public String next() throws IOException {
	        if(st.hasMoreTokens())
	            return st.nextToken();
	        else
	            st = new StringTokenizer(br.readLine());
	        return next();
	    }

	    public int nextInt() throws IOException {
	        return Integer.parseInt(next());
	    }
	    //#
	    public long nextLong() throws IOException {
	        return Long.parseLong(next());
	    }
	    public double nextDouble() throws IOException {
	        return Double.parseDouble(next());
	    }
	    //$
	}
}