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.ArrayDeque;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Deque;
import java.util.StringTokenizer;

public class TaskE {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		FastReader sc = new FastReader();
		PrintWriter out = new PrintWriter(System.out);
		
		int test = sc.nextInt();
		while(test-->0) {
			int n = sc.nextInt();
			
			int[] a = new int[n];
			for(int i =0;i<n;i++)
				a[i] = sc.nextInt();
			
			if(n==1) {
				out.println(1);
				out.println(1);
				continue;
			}
			ArrayList<Integer> al = new ArrayList<>();
			for(int i = 0;i<n;i++) {
				if(Arrays.binarySearch(a, i+1)<0) {
					al.add(i+1);
				}
			}
			Deque<Integer> deq = new ArrayDeque<>();
			for(int i = 0;i<n;i++) {
				if(Arrays.binarySearch(a, i+1)<0) {
					deq.add(i+1);
				}
			}
			
			for(int i = 0;i<n;i++) {
				if(i==0 || a[i]!=a[i-1]) {
					out.print(a[i]+" ");
				} else {
					out.print(deq.poll()+" ");
				}
			}
			out.println();
			int count = 0;
			
			long[] n1 = new long[n];
			n1[0] = a[0];
			for(int i = 0;i<n;i++) {
				if(i==0 || a[i]!=a[i-1]) {
					out.print(a[i]+" ");
				} else {
					int k = 0;
					for(int b = al.size();b>=1;b/=2) {
						while(k+b<al.size() && al.get(k+b)<a[i]) k+=b;
					}
					out.print(al.get(k)+" ");
					al.remove(k);
				}
			}
			out.println();
		}
		out.close();
	}
	static class FastReader {
		
		StringTokenizer st;
		BufferedReader br;
		
		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 s = "";
			while(st==null || st.hasMoreElements()) {
				try {
					s = br.readLine();
				} catch (IOException e) {
					e.printStackTrace();
				}
			}
			return s;
		}
	}
}