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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.TreeSet;
import java.util.StringTokenizer;
import java.io.BufferedReader;
import java.io.FileReader;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author Khater
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Scanner in = new Scanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        ERestoringThePermutation solver = new ERestoringThePermutation();
        solver.solve(1, in, out);
        out.close();
    }

    static class ERestoringThePermutation {
        public void solve(int testNumber, Scanner sc, PrintWriter pw) {
            int t = 1;
            t = sc.nextInt();
            while (t-- > 0) {
                int n = sc.nextInt();
                int[] arr = sc.nextIntArr(n);
                TreeSet<Integer> all = new TreeSet<>();
                TreeSet<Integer> all1 = new TreeSet<>();
                for (int i = 1; i <= n; i++) all.add(i);
                for (int i = 1; i <= n; i++) all1.add(i);
                int[] ans = new int[n];
                int[] ans1 = new int[n];
                Arrays.fill(ans, -1);
                Arrays.fill(ans1, -1);
                for (int i = 0; i < n; i++) {
                    if (all.contains(arr[i])) {
                        ans[i] = arr[i];
                        ans1[i] = arr[i];
                        all.remove(arr[i]);
                        all1.remove(arr[i]);
                    }
                }
                int max = 0;
                for (int i = 0; i < n; i++) {
                    max = Math.max(max, arr[i]);
                    if (ans[i] == -1) {
                        ans[i] = all.floor(max);
                        all.remove(ans[i]);
                    }
                }
                max = 0;
                for (int i = n - 1; i >= 0; i--) {
                    max = Math.max(max, arr[i]);
                    if (ans1[i] == -1) {
                        ans1[i] = all1.floor(max);
                        all1.remove(ans1[i]);
                    }
                }
                for (int x : ans1) pw.print(x + " ");
                pw.println();
                for (int x : ans) pw.print(x + " ");
                pw.println();
            }

        }

    }

    static class Scanner {
        StringTokenizer st;
        BufferedReader br;

        public Scanner(FileReader r) {
            br = new BufferedReader(r);
        }

        public Scanner(InputStream s) {
            br = new BufferedReader(new InputStreamReader(s));
        }

        public String next() {
            while (st == null || !st.hasMoreTokens()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return st.nextToken();
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }

        public int[] nextIntArr(int n) {
            try {
                int[] a = new int[n];
                for (int i = 0; i < n; i++)
                    a[i] = nextInt();
                return a;
            } catch (Exception e) {
                throw new RuntimeException(e);
            }

        }

    }
}