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.lang.reflect.Array;
import java.util.*;

public class Main {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int t = scanner.nextInt();
        int[] mass = new int[200_001];
        int[] ans = new int[200_001];
        int[] bigAns = new int[200_001];
        NavigableSet<Integer> set = new TreeSet<>();
        NavigableSet<Integer> ff = new TreeSet<>();
        StringBuilder stringBuilder = new StringBuilder();
        for (int z = 0 ; z < t ; ++z) {
            int n = scanner.nextInt();
            set.clear();
            for (int i = 0 ; i < n ; ++i) {
                ans[i] = -1;
                bigAns[i] = -1;
                mass[i] = scanner.nextInt();
                set.add(i + 1);
                ff.add(i + 1);
            }
            ans[0] = bigAns[0] = mass[0];
            set.remove(ans[0]);
            ff.remove(ans[0]);
            for (int i = 1 ; i < n ; ++i) {
                if (mass[i] != mass[i - 1]) {
                    ans[i] = bigAns[i] = mass[i];
                    set.remove(ans[i]);
                    ff.remove(ans[i]);
                }
            }
            for (int i = 0 ; i < n ; ++i) {
                if (ans[i] == -1) {
                    int min = set.pollFirst();
                    ans[i] = min;
                }
                stringBuilder.append(ans[i]).append(' ');
            }
            stringBuilder.append('\n');
            for (int i = 0 ; i < n ; ++i) {
                if (bigAns[i] == -1) {
                    int kinch = bigAns[i - 1];
                    for (int j = i ; j < n ; ++j) {
                        if (bigAns[j] != -1) {
                            i = j - 1;
                            break;
                        }
                        int curPizda = ff.lower(kinch);
                        ff.remove(curPizda);
                        bigAns[j] = curPizda;
                    }
                }
            }
            for (int i = 0 ; i < n ; ++i) stringBuilder.append(bigAns[i]).append(' ');
            stringBuilder.append('\n');

        }
        System.out.println(stringBuilder);
    }

}