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 Codeforces {

    static FastScanner sc;
    static PrintWriter out;

    public static void main(String[] args){
        sc = new FastScanner();
        out = new PrintWriter(System.out);

        int t = sc.nextInt();

        while(t-- > 0){
            solve();
        }

        out.close();
    }

    private static void solve() {

        int n = sc.nextInt();
        int[] q = sc.readArray(n);

        PriorityQueue<Integer> q1 = new PriorityQueue<>();
        PriorityQueue<Integer> q2 = new PriorityQueue<>(Collections.reverseOrder());

        int[] op1 = new int[n];
        int[] op2 = new int[n];


        op1[0] = q[0];
        op2[0] = q[0];
        for(int i=1; i<q[0]; i++){
            q1.add(i);
            q2.add(i);
        }
        int prev = q[0] + 1;
        for(int i=1; i<n; i++){
            if(q[i] > q[i-1]){
                op1[i] = q[i];
                op2[i] = q[i];
                for(int j=prev; j<q[i]; j++){
                    q1.add(j);
                    q2.add(j);
                }
                prev = q[i] + 1;
            }
            else {
                op1[i] = q1.poll();
                op2[i] = q2.poll();
            }
        }
        for(int num: op1)
            out.print(num + " ");
        out.println();
        for(int num: op2)
            out.print(num + " " );
        out.println();
    }

    private static int findLcm(int n, int m) {

        return n*m/findGcd(Math.min(n, m), Math.max(n, m));
    }

    private static int findGcd(int n, int m) {
        if(n == 0)
            return m;
        return findGcd(m%n, n);
    }

    public static void intSort (int[] arr) {
        //First ruffle

        int n = arr.length;
        Random r = new Random();
        for (int i=0; i<arr.length; i++) {
            int j = r.nextInt(n);
            int temp = arr[i];
            arr[i] = arr[j];
            arr[j] = temp;
        }

        //then sort
        Arrays.sort(arr);
    }

    public static void longSort (long[] arr) {
        //First ruffle

        int n = arr.length;
        Random r = new Random();
        for (int i=0; i<arr.length; i++) {
            int j = r.nextInt(n);
            long temp = arr[i];
            arr[i] = arr[j];
            arr[j] = temp;
        }

        //then sort
        Arrays.sort(arr);
    }

    public static void sort2dArray (long[][] arr) {
        Arrays.sort(arr, (a, b) -> {
            if (a[1] == b[1])
                return (int) (a[0] - b[0]);
            return (int) (a[1] - b[1]);
        });
    }

    public static class FastScanner {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st=new StringTokenizer("");
        String next() {
            while (!st.hasMoreTokens())
                try {
                    st=new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }
        int[] readArray(int n) {
            int[] a=new int[n];
            for (int i=0; i<n; i++) a[i]=nextInt();
            return a;
        }
        long nextLong() {
            return Long.parseLong(next());
        }
    }

}