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.StringTokenizer;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author koneko096
 */
public class Main {
  public static void main(String[] args) {
    InputStream inputStream = System.in;
    OutputStream outputStream = System.out;
    InputReader in = new InputReader(inputStream);
    PrintWriter out = new PrintWriter(outputStream);
    ERestoringThePermutation solver = new ERestoringThePermutation();
    int testCount = Integer.parseInt(in.next());
    for (int i = 1; i <= testCount; i++)
      solver.solve(i, in, out);
    out.close();
  }

  static class ERestoringThePermutation {
    private int[] prv;
    private int[] nxt;

    public void solve(int testNumber, InputReader in, PrintWriter out) {
      int n = in.nextInt();

      prv = new int[n + 4];
      nxt = new int[n + 4];
      for (int i = 0; i <= n + 1; i++) {
        prv[i] = i - 1;
        nxt[i] = i + 1;
      }

      int[] A = new int[n + 2];
      int[] ans1 = new int[n + 2];
      int[] ans2 = new int[n + 2];
      boolean[] mark = new boolean[n + 4];
      for (int i = 1; i <= n; i++) {
        int x = in.nextInt();
        A[i] = x;
        if (A[i] != A[i - 1]) {
          mark[x] = true;
          ans2[i] = x;
          ans1[i] = x;
        }
      }

      int cur = 1;
      for (int i = 1; i <= n; i++) {
        if (ans1[i] != 0) continue;
        while (mark[cur]) cur++;
        ans1[i] = cur;
        mark[cur] = true;
      }

      cur = 0;
      for (int i = 1; i <= n; i++) {
        if (ans2[i] != 0) {
          cur = prv[ans2[i]];
          updateLink(ans2[i]);
          continue;
        }
        ans2[i] = cur;
        updateLink(cur);
        cur = prv[cur];
      }

      for (int i = 1; i <= n; i++) {
        out.print(ans1[i] + " ");
      }
      out.println();
      for (int i = 1; i <= n; i++) {
        out.print(ans2[i] + " ");
      }
      out.println();
    }

    private void updateLink(int cur) {
      int pr = prv[cur], ne = nxt[cur];
      nxt[pr] = ne;
      prv[ne] = pr;
    }

  }

  static class InputReader {
    private BufferedReader reader;
    private 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());
    }

  }
}