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:
// Oh boy this is gonna be fun

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;

public class E {

    static FastScanner fs=new FastScanner();

    public static void main(String[] args) {
        int tt = 1;
        tt = fs.nextInt();
        for(int w = 0; w < tt; w++) {
            solve();
        }
    }

    private static void solve() {
        int n = fs.nextInt();
        int[] a = fs.readArray(n);
        for (int i = 0; i < n; i++)
            --a[i];
        goMin(a);
        goMax(a);
    }

    private static void goMin(int[] a) {
        int n = a.length;
        boolean[] used = new boolean[n];
        int nxt = 0;
        int last = -1;
        int[] ans = new int[n];
        for(int i = 0; i < n; i++) {
            if(a[i] > last) {
                last = a[i];
                ans[i] = last;
            } else {
                while(used[nxt])
                    ++nxt;
                ans[i] = nxt;
            }
            used[ans[i]] = true;
        }
        for(int i = 0; i < n; i++) {
            System.out.print(ans[i]+1 + " ");
        }
        System.out.println();
    }

    private static void goMax(int[] a) {
        int n = a.length;
        int[] ans = new int[n];
        int last = -1;
        PriorityQueue<Integer> pq = new PriorityQueue<>(n, Collections.reverseOrder());
        for(int i = 0; i < n; i++) {
            if(a[i] > last) {
                for(int x = last + 1; x < a[i]; x++) {
                    pq.offer(x);
                }
                last = a[i];
                ans[i] = last;
            } else {
                int x = pq.poll();
                ans[i] = x;
            }
        }

        for(int i = 0; i < n; i++) {
            System.out.print(ans[i]+1 + " ");
        }
        System.out.println();
    }

    private static void mySort(int[] a) {
        int n = a.length;
        Random rd = new Random(420);
        for (int i = 0; i < n; i++) {
            int j = rd.nextInt(i+1);
            int tmp = a[j];
            a[j] = a[i];
            a[i] = tmp;
        }
        Arrays.sort(a);
    }

    static class FastScanner {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st=new StringTokenizer("");
        public String next() {
            while (!st.hasMoreElements())
                try {
                    st=new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            return st.nextToken();
        }

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

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

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

}