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.*;

public class SolutionE extends Thread {
    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;
        }
    }

    private static final FastReader scanner = new FastReader();
    private static final PrintWriter out = new PrintWriter(System.out);

    public static void main(String[] args) {
        new Thread(null, new SolutionE(), "Main", 1 << 26).start();
    }

    public void run() {
        int t = scanner.nextInt();
        for (int i = 0; i < t; i++) {
            solve();
        }
        out.close();
    }

    static class Node {
        int number;
        Node prev;
        Node next;

        public Node(int number) {
            this.number = number;
        }
    }

    private static void solve() {
        int n = scanner.nextInt();

        int[] q = new int[n];
        for (int i = 0; i < n; i++) {
            q[i] = scanner.nextInt();
        }



        boolean[] used = new boolean[n+1];
        int pointerUsed = 1;
        int lastNumberSeen = -1;
        int[] pMin = new int[n];
        for (int i = 0; i < n; i++) {
            if (lastNumberSeen != q[i]){
                pMin[i] = q[i];
                used[q[i]] = true;
            } else {
                while (used[pointerUsed]) {
                    pointerUsed++;
                }
                pMin[i] = pointerUsed;
                used[pointerUsed] = true;
            }

            lastNumberSeen = q[i];
        }
        StringBuilder s = new StringBuilder();
        for (int i = 0; i < n; i++) {
            s.append(pMin[i])
             .append(" ");
        }
        out.println(s);

        Node[] nodes = new Node[n+2];
        for (int i = 0; i <= n+1; i++) {
            nodes[i] = new Node(i);
            if (i >= 1) {
                nodes[i].next = nodes[i-1];
            }
        }
        for (int i = 0; i <= n; i++) {
            nodes[i].prev = nodes[i+1];
        }

        int maxNumberSeen = 0;
        lastNumberSeen = 0;
        StringBuilder t = new StringBuilder();
        Node pointerNodes = null;

        for (int i = 0; i < n; i++) {
            if (lastNumberSeen != q[i]) {
                t.append(q[i])
                 .append(" ");
                Node prev = nodes[q[i]].prev;
                Node next = nodes[q[i]].next;
                prev.next = next;
                next.prev = prev;
                if (q[i] > maxNumberSeen) {
                    maxNumberSeen = q[i];
                    pointerNodes = nodes[q[i]].next;
                }
            } else {
                t.append(pointerNodes.number)
                 .append(" ");
                Node prev = pointerNodes.prev;
                Node next = pointerNodes.next;
                prev.next = next;
                next.prev = prev;
                pointerNodes = pointerNodes.next;
            }

            lastNumberSeen = q[i];
        }
        out.println(t);
    }

    //REMINDERS:
    //- CHECK FOR INTEGER-OVERFLOW BEFORE SUBMITTING

    //- CAN U BRUTEFORCE OVER SOMETHING, TO MAKE IT EASIER TO CALCULATE THE SOLUTION
}