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.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.StringTokenizer;
import java.util.TreeSet;

public class problemE {
    static class Solution {
        void solve() {
            int n = fs.nextInt();
            int[] a = fs.readArray(n);
            smaller(a, n);
            larger(a, n);
        }

        void smaller(int[] a, int n) {
            int last = -1;
            int[] ans = new int[n];
            int ind = 0 ;

            TreeSet<Integer> set = new TreeSet<>();
            for (int i = 1; i <= n ; i ++ ) set.add(i);

            for (int x : a) {
                if (x != last) {
                    ans[ind] = x;
                    last = x;
                    ind ++;
                    set.remove(x);
                } else {
                    ans[ind] = set.first();
                    set.remove(ans[ind]);
                    ind++;
                }
            }
            out.printArray(ans);
        }

        void larger(int[] a, int n) {
            int last = -1;
            int[] ans = new int[n];
            int ind = 0 ;

            TreeSet<Integer> set = new TreeSet<>();
            for (int i = 1; i <= n ; i ++ ) set.add(i);

            for (int x : a) {
                if (x != last) {
                    ans[ind] = x;
                    last = x;
                    ind ++;
                    set.remove(x);
                } else {
                    ans[ind] = set.floor(x);
                    set.remove(ans[ind]);
                    ind++;
                }
            }
            out.printArray(ans);

        }
    }

    public static void main(String[] args) throws Exception {
        int T = 1;
        T = fs.nextInt();
        Solution solution = new Solution();
        for (int t = 0; t < T; t++) solution.solve();
        out.close();
    }

    static void debug(Object... O) {
        System.err.println("DEBUG: " + Arrays.deepToString(O));
    }

    private static FastScanner fs = new FastScanner();
    private static Output out = new Output();

    static class Output extends PrintWriter {
        Output() {
            super(System.out);
        }

        void printArray(int[] a) {
            for (int x : a ) { print(x); print(' '); }
            println();
        }

        void printArray(long[] a) {
            for (long x : a ) { print(x); print(' '); }
            println();
        }

        void printAll(Object ...a) {
            for (Object o : a) {print(o.toString()); print(' ');}
            println();
        }

//        void printAll(int ...a) {
//            for (int x : a) {print(x); print(' ');}
//            println();
//        }
//
//        void printAll(long ...a) {
//            for (long x : a) {print(x); print(' ');}
//            println();
//        }
    }

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

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

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

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

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

        ArrayList<Integer> readArrayList(int n) {
            ArrayList<Integer> a = new ArrayList<>(n);
            for (int i = 0 ; i < n; i ++ ) a.add(fs.nextInt());
            return a;
        }

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

        double nextDouble() {
            return Double.parseDouble(next());
        }

        String nextString() {
            return next();
        }
    }
}