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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.HashSet;
import java.util.PriorityQueue;
import java.util.StringTokenizer;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.InputStream;
import java.util.TreeSet;

/*
@author kalZor
 */
public class TaskE {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        FastReader in = new FastReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        Solver solver = new Solver();
        int testCount = 1;
        testCount = Integer.parseInt(in.next());
        for (int i = 1; i <= testCount; i++)
            solver.solve(i, in, out);
        out.close();
    }

    static class Solver {
        public void print(int a,int b,PrintWriter out){
            out.println("? "+a+" "+b);
            out.flush();
            return;
        }
        public void solve(int testNumber, FastReader in, PrintWriter out) {
            int n = in.nextInt();
            int[] a = in.nextArray(n);
            int[] ansMin = new int[n];
            int[] ansMax = new int[n];
            PriorityQueue<Integer> pq = new PriorityQueue<>();
            TreeSet<Integer> ts = new TreeSet<>();
            HashSet<Integer> set = new HashSet<>();
            ansMax[0]  =a[0];
            ansMin[0]  =a[0];
            set.add(a[0]);
            for(int i=1;i<n;i++){
                if(a[i]==a[i-1]){
                    ansMax[i] = -1;
                    ansMin[i] = -1;
                }
                else{
                    ansMax[i] = a[i];
                    ansMin[i] = a[i];
                    set.add(a[i]);
                }
            }
            for(int i=1;i<n;i++){
                if(!set.contains(i)){
                    pq.add(i);
                    ts.add(i);
                }
            }
            for(int i=0;i<n;i++){
                if(ansMin[i]==-1){
                    ansMin[i] = pq.poll();
                }
            }
            int ini = 0;
            for(int i=0;i<n;i++){
                if(ansMax[i]==-1) {
                    ansMax[i] = ts.lower(ini);
                    ts.remove(ansMax[i]);
                }
                ini = ansMax[i];
            }
            for(int x:ansMin) out.print(x+" ");
            out.println();
            for(int x:ansMax) out.print(x+" ");
            out.println();
        }
    }

    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

        public FastReader(InputStream stream){
            br = new BufferedReader(new
                InputStreamReader(stream));
        }

        public String next(){
            while (st == null || !st.hasMoreElements()){
                try{
                    st = new StringTokenizer(br.readLine());
                }
                catch (IOException e){
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        public int nextInt(){
            return Integer.parseInt(next());
        }

        public long nextLong(){
            return Long.parseLong(next());
        }

        public double nextDouble(){
            return Double.parseDouble(next());
        }

        public String nextLine(){
            String str = "";
            try{
                str = br.readLine();
            }
            catch (IOException e){
                e.printStackTrace();
            }
            return str;
        }

        public int[] nextArray(int n) {
            int[] a=new int[n];
            for (int i=0; i<n; i++) a[i]=nextInt();
                return a;
        }
    }
}