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.util.*;
import java.io.*;

public class Solution
{
    static class Reader {
		BufferedReader br;
		StringTokenizer st;
 
		public Reader() {
			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[] nextArr(int n)
		{
		    int a[]=new int[n];
		    for (int i=0;i<n;i++)a[i]=nextInt();
		    return a;
		}
		
		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;
		}
 
	}
    static class Ele implements Comparable<Ele> 
    {  
        public int x,y;  
        Ele(int x1,int y1) 
        {  
            x=x1;y=y1; 
        }  
        public int compareTo(Ele ob) {  
        if(ob.x!=x)return x-ob.x;
        return this.y-ob.y;  
        }   
        public String toString()
        {
        	return "["+x+","+y+"]";
        }
    }
	void disp(PrintWriter o,boolean b)
    {
        if (b) o.println("Yes");
        else o.println("No");
    }
    void disp(PrintWriter o,int ...a)
    {
        o.println(Arrays.toString(a));
    }
    void disp(PrintWriter o,long ...a)
    {
        o.println(Arrays.toString(a));
    }
    int dp[][];

	public static void main(String[] args) throws IOException 
	{
		Reader sc=new Reader();Solution G=new Solution();
		PrintWriter o = new PrintWriter(System.out);
		int t=1;t=sc.nextInt();
		int mod=(int)1e9+7;
		int x,x0,x1,x2;int y,y0,y1,y2;int s,s0,s1,s2;
		int n,m;int a[],b[],in[],in1[];
		long k,l;boolean v[],b1,b2;String ss;char c1[],c2[];
		//long l;long a[]; 
		ArrayList<ArrayList<Integer>> ll=new ArrayList<>();
		ArrayList<Integer> a1=new ArrayList<>();
		ArrayList<Integer> a2=new ArrayList<>();
		PriorityQueue<Integer> pq1=new PriorityQueue<>();
		PriorityQueue<Integer> pq2=new PriorityQueue<>(Collections.reverseOrder());
		ArrayDeque<Integer> dq=new ArrayDeque<>();
		TreeSet<Integer> h0=new TreeSet<>();
		TreeSet<Integer> h1=new TreeSet<>();
		TreeMap<Integer,Integer> h=new TreeMap<>();
		try{
		while (t-->0)
		{
		    n=sc.nextInt();a=sc.nextArr(n);b=new int[n];
		    Arrays.fill(b,-1);
		    for (int i=1;i<=n;i++){h0.add(i);h1.add(i);}
		    b[0]=a[0];h0.remove(b[0]);h1.remove(b[0]);
		    for (int i=1;i<n;i++)if (a[i]!=a[i-1]){b[i]=a[i];h0.remove(b[i]);h1.remove(b[i]);}
		    //o.println(Arrays.toString(b));
		    int cm=a[0];
		    for (int i=0;i<n;i++)
		    {
		        
		        if (b[i]==-1)
		        {
		            //o.println(h0.floor(b[i]));o.println(h1.ceiling(b[i]));
		            
		            a1.add(h0.pollFirst());a2.add(h1.floor(cm));
		            h1.remove(h1.floor(cm));
		        }
		        else
		        {
		            a1.add(b[i]);a2.add(b[i]);
		            cm=b[i];
		        }
		        //o.println(h0);
		        //o.println(h1);
		    }
		    
		    for (Integer i:a1)o.print(i+" ");
		    o.println();
		    for (Integer i:a2)o.print(i+" ");
		    o.println();
		    //o.println(a2);
		    //o.println();
		    //o.println();
		    //o.println(h);
		    //o.println(x2);
		    //o.println();
		    h0.clear();ll.clear();a1.clear();a2.clear();h1.clear();h.clear();pq1.clear();pq2.clear();
		}
		}
		catch (Throwable e)
		{
		    e.printStackTrace();
		}
		//o.println("HI");
		
        o.flush();
        o.close();
	}
}