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 static java.lang.Math.max;
import static java.lang.Math.min;
import static java.lang.Math.abs;
import java.io.*;
import java.util.*;

public class B
{
    static int n;
    static int[] arr;
    static char[] s;
    public static void main(String[] args) throws IOException
    {
        f = new Flash();
        out = new PrintWriter(System.out);
        
        int T = ni();
        for(int tc = 1; tc <= T; tc++){
            n = ni(); arr = arr(n); fn();
        }
        
        out.flush(); out.close();
    }
    
    static void fn()
    {
        TreeSet<Integer> set = new TreeSet<>();
        TreeSet<Integer> set2 = new TreeSet<>();
        for(int i = 1; i <= n; i++) {
        	set.add(i); set2.add(i);
        }
        
        int[] min = new int[n], max = new int[n];
        min[0] = arr[0]; max[0] = arr[0];
        set.remove(arr[0]); set2.remove(arr[0]);
        for(int i = 1; i < n; i++) {
        	if(arr[i] == arr[i-1]) continue;
        	min[i] = max[i] = arr[i];
        	set.remove(arr[i]);
        	set2.remove(arr[i]);
        }
        
        for(int i = 1; i < n; i++) {
        	if(arr[i] != arr[i-1]) continue;
        	min[i] = set2.ceiling(-1);
        	set2.remove(min[i]);
        	max[i] = set.floor(arr[i]-1);
        	set.remove(max[i]);
        }
        
        print(min); print(max);
    }
    
    static Flash f;
    static PrintWriter out;
    static final long mod = (long)1e9+7;
    static final long inf = Long.MAX_VALUE;
    static final int _inf = Integer.MAX_VALUE;
    static final int maxN = (int)5e5+5;
    static long[] fact, inv;
    
    static void sort(int[] a){
        List<Integer> A = new ArrayList<>();
        for(int i : a) A.add(i);
        Collections.sort(A);
        for(int i = 0; i < A.size(); i++) a[i] = A.get(i);
    }
    
    static void sort(long[] a){
        List<Long> A = new ArrayList<>();
        for(long i : a) A.add(i);
        Collections.sort(A);
        for(int i = 0; i < A.size(); i++) a[i] = A.get(i);
    }
    
    static void print(int[] a){
        StringBuilder sb = new StringBuilder();
        for(int i = 0; i < a.length; i++) sb.append(a[i] + " ");
        sop(sb);
    }
    
    static void print(long[] a){
        StringBuilder sb = new StringBuilder();
        for(int i = 0; i < a.length; i++) sb.append(a[i] + " ");
        sop(sb);
    }
    
    static int swap(int itself, int dummy){return itself;}
    static long swap(long itself, long dummy){return itself;}
    static void sop(Object o){out.println(o);}
    static int ni(){return f.ni();}
    static long nl(){return f.nl();}
    static double nd(){return f.nd();}
    static String next(){return f.next();}
    static String ns(){return f.ns();}
    static char[] nc(){return f.nc();}
    static int[] arr(int len){return f.arr(len);}
    static int gcd(int a, int b){if(b == 0) return a; return gcd(b, a%b);}
    static long gcd(long a, long b){if(b == 0) return a; return gcd(b, a%b);}
    static int lcm(int a, int b){return (a*b)/gcd(a, b);}
    static long lcm(long a, long b){return (a*b)/gcd(a, b);}
    
    static class Flash
    {
        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();
        }
        
        String ns(){
            String s = new String();
            try{
                s = br.readLine().trim();
            }catch(IOException e){
                e.printStackTrace();
            }
            
            return s;
        }
        
        int[] arr(int n){
            int[] a = new int[n];
            for(int i = 0; i < n; i++) a[i] = ni();
            return a;
        }
        
        char[] nc(){return ns().toCharArray();}
        int ni(){return Integer.parseInt(next());}
        long nl(){return Long.parseLong(next());}
        double nd(){return Double.parseDouble(next());}
    }
}