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.*;
import java.lang.Math;
public class a {
    InputStream inputStream = System.in;
    OutputStream outputStream = System.out;
    InputReader in;
    PrintWriter out;
    public static void main(String[] arg) {
        new a().run();
    }

    public void solve() throws IOException {
        int n = in.nextInt();
        int[] A = new int[n + 1], maxA = new int[n + 1], minA = new int[n + 1];
        TreeSet<Integer> maxi = new TreeSet<>(), mini = new TreeSet<>();
        int l = 0;
        A[0] = 0;
        for (int i = 1; i <= n; ++i) {
            A[i] = in.nextInt();
            for (int j = l + 1; j <= A[i]; ++j) {
                maxi.add(j);
                mini.add(j);
            }
            l = A[i];
            if (A[i] != A[i - 1]) {
                maxA[i] = A[i];
                minA[i] = A[i];
            } else {
                maxA[i] = maxi.last();
                minA[i] = mini.first();
            }
            maxi.remove(maxA[i]);
            mini.remove(minA[i]);
        }
        for (int i = 1; i <= n; ++i) {
            out.print(minA[i] + " ");
        }
        out.println();
        for (int i = 1; i <= n; ++i) {
            out.print(maxA[i] + " ");
        }
        out.println();
    }

    public void run() {
        try {
            in = new InputReader(inputStream);
            out = new PrintWriter(outputStream);
            int t = 1;
            t = in.nextInt();
            while (t-- > 0) {
                solve();
            }            
            out.close();
        } catch (IOException e) {
            e.printStackTrace();
        }
    }

    static class InputReader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;
 
        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }
 
        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }
 
        public int nextInt() {
            return Integer.parseInt(next());
        }
    }
}