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 CPP):

Solution:
#include <iostream>
#include <vector>
#include <queue>
#include <cstring>
#include <map>;
#include <set>;
#include <algorithm>

using namespace std;

int main()
{
    int t;
    cin >> t;
    int n;
    for(int i = 0; i < t; i++)
    {
        cin >> n;
        int mise[n];
        int pos[n];
        memset(pos, 0, sizeof(pos));
        map<int, int> T;
        set<int> comp;
        priority_queue<int, vector<int>, greater<int> > L;

        for(int i = 0; i < n; i++)
        {
            cin >> mise[i];
            if(T[mise[i]] == 0)
            {
                pos[i] = mise[i];
            }
            T[mise[i]] = 1;
            comp.insert(i + 1);
        }
        for(int i = 1; i <= n; i++)
        {
            if(T[i] == 0)
            {
                L.push(i);
            }
        }

        for(int i = 0; i < n; i++)
        {
            if(pos[i] == 0)
            {
                pos[i] = L.top();
                L.pop();
            }
        }

        for(int i = 0; i < n; i++)
        {
            if(i == 0)
            {
                cout << pos[i];
            }
            else
            {
                cout << " " << pos[i];
            }
        }
        cout << "\n";
        int pre;

        for(int i = 0; i < n; i++)
        {
            if(T[pos[i]] == 1)
            {
                pre = pos[i];
                comp.erase(pos[i]);
                continue;
            }
            else
            {
                pos[i] = *(--comp.upper_bound(pre));
                comp.erase(pos[i]);
            }
        }
        for(int i = 0; i < n; i++)
        {
            if(i == 0)
            {
                cout << pos[i];
            }
            else
            {
                cout << " " << pos[i];
            }
        }
        cout << "\n";
    }
    return 0;
}