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.util.StringTokenizer;

public class E_Restoring_the_Permutation {

     static class FastReader {
          BufferedReader br;
          StringTokenizer st;

          public FastReader() {
               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 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;
          }
     }
     public static void main(String[] args) {
          FastReader sc = new FastReader();
          int[] a = new int[200003];
          boolean[] b = new boolean[200003];
          int[] c = new int[200003];
          int t = sc.nextInt();
          while (t > 0) {
               t --;
               int n = sc.nextInt();
               for (int i = 1; i <= n; ++ i) {
                    a[i] = sc.nextInt();
                    b[i] = true;
               }
               int p = 1;
               for (int i = 1; i <= n; ++ i) {
                    int tmp = 0;
                    if (i == 1 || a[i] != a[i - 1]) {
                         tmp = a[i];
                    }
                    else {
                         while (! b[p]) {
                              ++ p;
                         }
                         tmp = p;
                    }
                    b[tmp] = false;
                    System.out.print(tmp + " ");
               }
               System.out.println("");
               p = 0;
               int m = 0;
               for (int i = 1; i <= n; ++ i) {
                         while (p < a[i]) {
                              ++ p;
                              ++ m;
                              c[m] = p;
                         }
                    System.out.print(c[m] + " ");
                    m --;
               }
               System.out.println("");
          }
     }
}