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:
/* package codechef; // don't place package name! */

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

/* Name of the class has to be "Main" only if the class is public. */
public class Codechef
{
	public static void main (String[] args) throws java.lang.Exception
	{
		Reader sc = new Reader();
		BufferedWriter out = new BufferedWriter(new OutputStreamWriter(System.out));
		int t = sc.nextInt();
		
		while(t-- > 0) {
		   int n = sc.nextInt();
		   int[] p = new int[n];
		   for(int i = 0; i < n; i++) {
		       p[i] = sc.nextInt();
		   }
		   
		   int[] a = new int[n];
		   int[] b = new int[n];
		   Set<Integer> set = new HashSet<>();
		   for(int i = 0; i < n; i++) {
		       if(i == 0 || p[i-1] != p[i]) {
		           a[i] = p[i];
		           b[i] = p[i];
		       }
		       set.add(p[i]);
		   }
		   TreeSet<Integer> ts = new TreeSet<>();
		   PriorityQueue<Integer> minpq = new PriorityQueue<>(n);
		   
		   for(int i = 1; i <= n; i++) {
		       if(!set.contains(i)) {
		        minpq.add(i);
		        ts.add(i);
		       }
		   }
		   
		   for(int i = 0; i < n; i++) {
		       if(a[i] == 0) {
		           a[i] = minpq.poll();
		           b[i] = ts.floor(p[i]);
		           ts.remove(b[i]);
		       }
		   }
		   
		   for(int i = 0; i < n; i++) {
		       out.write(a[i]+" ");
		   }
		   out.write("\n");
		   for(int i = 0; i < n; i++) {
		       out.write(b[i]+" ");
		   }
		   out.write("\n");
		   out.flush();
		   
		}
		out.close();
	}
	
	static class Reader {
        final private int BUFFER_SIZE = 1 << 16;
        private DataInputStream din;
        private byte[] buffer;
        private int bufferPointer, bytesRead;
 
        public Reader()
        {
            din = new DataInputStream(System.in);
            buffer = new byte[BUFFER_SIZE];
            bufferPointer = bytesRead = 0;
        }
 
        public Reader(String file_name) throws IOException
        {
            din = new DataInputStream(
                new FileInputStream(file_name));
            buffer = new byte[BUFFER_SIZE];
            bufferPointer = bytesRead = 0;
        }
 
        public String readLine() throws IOException
        {
            byte[] buf = new byte[64]; // line length
            int cnt = 0, c;
            while ((c = read()) != -1) {
                if (c == '\n') {
                    if (cnt != 0) {
                        break;
                    }
                    else {
                        continue;
                    }
                }
                buf[cnt++] = (byte)c;
            }
            return new String(buf, 0, cnt);
        }
 
        public int nextInt() throws IOException
        {
            int ret = 0;
            byte c = read();
            while (c <= ' ') {
                c = read();
            }
            boolean neg = (c == '-');
            if (neg)
                c = read();
            do {
                ret = ret * 10 + c - '0';
            } while ((c = read()) >= '0' && c <= '9');
 
            if (neg)
                return -ret;
            return ret;
        }
 
        public long nextLong() throws IOException
        {
            long ret = 0;
            byte c = read();
            while (c <= ' ')
                c = read();
            boolean neg = (c == '-');
            if (neg)
                c = read();
            do {
                ret = ret * 10 + c - '0';
            } while ((c = read()) >= '0' && c <= '9');
            if (neg)
                return -ret;
            return ret;
        }
 
        public double nextDouble() throws IOException
        {
            double ret = 0, div = 1;
            byte c = read();
            while (c <= ' ')
                c = read();
            boolean neg = (c == '-');
            if (neg)
                c = read();
 
            do {
                ret = ret * 10 + c - '0';
            } while ((c = read()) >= '0' && c <= '9');
 
            if (c == '.') {
                while ((c = read()) >= '0' && c <= '9') {
                    ret += (c - '0') / (div *= 10);
                }
            }
 
            if (neg)
                return -ret;
            return ret;
        }
 
        private void fillBuffer() throws IOException
        {
            bytesRead = din.read(buffer, bufferPointer = 0,
                                 BUFFER_SIZE);
            if (bytesRead == -1)
                buffer[0] = -1;
        }
 
        private byte read() throws IOException
        {
            if (bufferPointer == bytesRead)
                fillBuffer();
            return buffer[bufferPointer++];
        }
 
        public void close() throws IOException
        {
            if (din == null)
                return;
            din.close();
        }
    }
}