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 E {

	static long mod = (long) (1e9 + 7);

	public static void main(String[] args) throws IOException {
		Scanner scn = new Scanner(System.in);
		PrintWriter out = new PrintWriter(System.out);
		StringBuilder sb = new StringBuilder();
		int T = scn.ni(), tcs = 0;
		E: while (tcs++ < T) {
			int n = scn.ni();
			int q[] = scn.nia(n);
			TreeSet<Integer> ts = new TreeSet<>();
			for (int i = 1; i <= n; i++)
				ts.add(i);
			int min[] = new int[n];
			for (int i = 0; i < n; i++) {
				min[i] = ts.floor(q[i]);
				ts.remove(min[i]);
			}
			ts = new TreeSet<>();
			for (int i = 1; i <= n; i++)
				ts.add(i);
			int max[] = new int[n];
			for (int i = 0; i < n; i++) {
				if (ts.contains(q[i])) {
					max[i] = q[i];
					ts.remove(max[i]);
				} else {
					max[i] = ts.pollFirst();
				}
			}
			for (int v : max)
				sb.append(v + " ");
			sb.append("\n");
			for (int v : min)
				sb.append(v + " ");
			sb.append("\n");
		}
		out.print(sb);
		out.close();
	}

	static class Scanner {

		StringTokenizer st;
		BufferedReader br;

		public Scanner(InputStream s) {
			br = new BufferedReader(new InputStreamReader(s));
		}

		public String next() throws IOException {
			while (st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}

		public int ni() throws IOException {
			return Integer.parseInt(next());
		}

		public long nl() throws IOException {
			return Long.parseLong(next());
		}

		public int[] nia(int n) throws IOException {
			int a[] = new int[n];
			String sa[] = br.readLine().split(" ");
			for (int i = 0; i < n; i++)
				a[i] = Integer.parseInt(sa[i]);
			return a;
		}

		public long[] nla(int n) throws IOException {
			long a[] = new long[n];
			String sa[] = br.readLine().split(" ");
			for (int i = 0; i < n; i++)
				a[i] = Long.parseLong(sa[i]);
			return a;
		}

		public void sort(int[] a) {
			ArrayList<Integer> l = new ArrayList<>();
			for (int v : a)
				l.add(v);
			Collections.sort(l);
			for (int i = 0; i < a.length; i++)
				a[i] = l.get(i);
		}

		public void sort(long[] a) {
			ArrayList<Long> l = new ArrayList<>();
			for (long v : a)
				l.add(v);
			Collections.sort(l);
			for (int i = 0; i < a.length; i++)
				a[i] = l.get(i);
		}
	}
}