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.Arrays;
import java.util.Random;
import java.util.StringTokenizer;
import java.util.TreeSet;

public final class E {

    public static void main(String[] args) {
        final FastScanner fs = new FastScanner();
        final int t = fs.nextInt();
        for (int test = 0; test < t; test++) {
            final int n = fs.nextInt();
            final int[] arr = new int[n];
            final TreeSet<Integer> ts1 = new TreeSet<>();
            final TreeSet<Integer> ts2 = new TreeSet<>();
            for (int i = 0; i < n; i++) {
                ts1.add(i + 1);
                ts2.add(i + 1);
                arr[i] = fs.nextInt();
            }
            final int[] min = new int[n];
            final int[] max = new int[n];
            min[0] = max[0] = arr[0];
            ts1.remove(arr[0]);
            ts2.remove(arr[0]);
            for (int i = 1; i < n; i++) {
                if (arr[i] != arr[i - 1]) {
                    ts1.remove(arr[i]);
                    ts2.remove(arr[i]);
                    min[i] = max[i] = arr[i];
                }
            }
            for (int i = 1; i < n; i++) {
                if (max[i] != 0) {
                    continue;
                }
                final int v = ts2.lower(max[i - 1]);
                min[i] = ts1.pollFirst();
                max[i] = v;
                ts2.remove(v);
            }
            StringBuilder sb1 = new StringBuilder();
            StringBuilder sb2 = new StringBuilder();
            for (int i = 0; i < n; i++) {
                sb1.append(min[i]);
                sb1.append(' ');
                sb2.append(max[i]);
                sb2.append(' ');
            }
            System.out.println(sb1);
            System.out.println(sb2);
        }
    }

    static final class Utils {
        private static class Shuffler {
            private static void shuffle(int[] x) {
                final Random r = new Random();

                for (int i = 0; i <= x.length - 2; i++) {
                    final int j = i + r.nextInt(x.length - i);
                    swap(x, i, j);
                }
            }

            private static void shuffle(long[] x) {
                final Random r = new Random();

                for (int i = 0; i <= x.length - 2; i++) {
                    final int j = i + r.nextInt(x.length - i);
                    swap(x, i, j);
                }
            }

            private static void swap(int[] x, int i, int j) {
                final int t = x[i];
                x[i] = x[j];
                x[j] = t;
            }

            private static void swap(long[] x, int i, int j) {
                final long t = x[i];
                x[i] = x[j];
                x[j] = t;
            }
        }

        public static void shuffleSort(int[] arr) {
            Shuffler.shuffle(arr);
            Arrays.sort(arr);
        }

        public static void shuffleSort(long[] arr) {
            Shuffler.shuffle(arr);
            Arrays.sort(arr);
        }

        private Utils() {}
    }

    static class FastScanner {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer("");

        private String next() {
            while (!st.hasMoreTokens()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    //noinspection CallToPrintStackTrace
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

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

        long nextLong() {
            return Long.parseLong(next());
        }

        int[] nextIntArray(int n) {
            final int[] a = new int[n];
            for (int i = 0; i < n; i++) { a[i] = nextInt(); }
            return a;
        }

        long[] nextLongArray(int n) {
            final long[] a = new long[n];
            for (int i = 0; i < n; i++) { a[i] = nextLong(); }
            return a;
        }
    }
}