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.*;

public class CodeForces extends Functions {
    public static void  solve(){
      int n = sc.nextInt();
      int[] arr = sc.setIntArray(n);
      PriorityQueue<Integer> minHeap = new PriorityQueue<>();
      PriorityQueue<Integer> maxHeap = new PriorityQueue<>(Collections.reverseOrder());
      int[] min = new int[n];
      int[] max = new int[n];
      min[0] = arr[0];
      max[0] = arr[0];
        int x = 1;
        while (x<arr[0]){
            maxHeap.add(x);
            minHeap.add(x);
            x++;
        }
          for(int i=1;i<n;i++){
              if(arr[i]==arr[i-1]){

                  min[i] = minHeap.poll();
                  max[i] = maxHeap.poll();
              }else {
                  x = arr[i-1]+1;
                  while (x<arr[i]){
                      maxHeap.add(x);
                      minHeap.add(x);
                      x++;
                  }

                  min[i] = arr[i];
                  max[i] = arr[i];
              }
          }
      for(int ele : min) System.out.print(ele+" ");
        System.out.println();
        for(int ele : max) System.out.print(ele+" ");
        System.out.println();


    }




    public static void main(String[] args) {
        int testCase = 1;
        testCase= sc.nextInt();
        while (testCase-->0){
            solve();
        }
        long end = System.currentTimeMillis();
//        System.out.println("time took in ms : "+(end-start));
    }
    static Scanner sc = new Scanner();


}

class Functions {
    public static int mod = (int)1e9+7;
    public static int INT_MAX = Integer.MAX_VALUE;
    public static int INT_MIN = Integer.MIN_VALUE;
    public static long LONG_MAX = Long.MAX_VALUE;
    public static long LONG_MIN = Long.MIN_VALUE;
    public static double DOUBLE_MAX = Double.MAX_VALUE;
    public static double DOUBLE_MIN = Double.MIN_VALUE;

    public static void sort(int[] a,boolean isAscending){
        ArrayList<Integer> temp = new ArrayList<>();
        for (int j : a) temp.add(j);
        sort(temp,isAscending);
        for(int i=0;i<a.length;i++)a[i] = temp.get(i);
    }
    public static void sort(List T,boolean isAscending){
        if(isAscending)
            Collections.sort(T);
        else Collections.sort(T,Collections.reverseOrder());
    }
    public static long factorial(int n){
        long fact = 1L;
        for(int i=2;i<=n;i++)fact = (fact*i)%mod;
        return fact;
    }

    public  static int ncr(int n, int r){
        // time O(n+r)
        if (r > n)
            return 0;
        long[] inv = new long[r + 1];
        inv[1] = 1;
        // Getting the modular inversion
        // for all the numbers
        // from 2 to r with respect to m
        for (int i = 2; i <= r; i++) {
            inv[i] = mod - (mod / i) * inv[(int) (mod % i)] % mod;
        }
        int ans = 1;
        // for 1/(r!) part
        for (int i = 2; i <= r; i++) {
            ans = (int) (((ans % mod) * (inv[i] % mod)) % mod);
        }
        // for (n)*(n-1)*(n-2)*...*(n-r+1) part
        for (int i = n; i >= (n - r + 1); i--) {
            ans = (int) (((ans % mod) * (i % mod)) % mod);
        }
        return ans;
    }

    public static void reverseArray(int[] a){
        int left = 0;
        int right = a.length-1;
        while (left<right){
            int temp =a[left];
            a[left] = a[right];
            a[right] = temp;
            left++;
            right--;
        }
    }
    public static long isPrime(int n){
        for(long i=2;i*i<=n;i++)
            if(n%i==0)return i;
        return -1;
    }
}

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

    }
    public int nextInt(){return Integer.parseInt(next());}
    public long nextLong(){return Long.parseLong(next());}
    public double nextDouble(){return Double.parseDouble(next());}
    public int[] setIntArray(int n){
        int[] arr =new int[n];
        for(int i=0;i<n;i++)arr[i] = nextInt();
        return arr;
    }
    public Integer[] setIntegerArray(int n){
        Integer[] arr =new Integer[n];
        for(int i=0;i<n;i++)arr[i] = nextInt();
        return arr;
    }
    public long[] setlongArray(int n){
        long[] arr =new long[n];
        for(int i=0;i<n;i++)arr[i] = nextLong();
        return arr;
    }
    public int[][] set2DIntegerMatrix(int row,int col){
        int[][] arr = new int[row][col];
        for(int i=0;i<row;i++)
            for(int j= 0;j<col;j++)
                arr[i][j] = nextInt();
        return arr;
    }
}