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:
/*
 * generated by github.com/calgagi/cp_gen
 * author: weiyutang
 * file: E.cpp
 * time: Fri Feb 19 14:35:44 2021
 */

#include <bits/stdc++.h>
using namespace std;
const string CP_GEN_FILE = "E";

namespace cp {

}

/* ===== BEGIN USER CODE ===== */

#define ll long long
#define ld long double
#define ii pair<int, int>
#define f first
#define s second
#define endl "\n"

void solve() {
    int n;
    cin >> n;
    vector<int> v(n);
    for (int& x: v) cin >> x;

    // O(n) Find all missing value smaller than n
    map<int, int> unused;
    for (int i = 0, val = 1; i < n; i++) {
        while (val < v[i]) {
            unused[val]++;
            val++;
        }
        if (val == v[i]) val++;
    }

    // O(n) No unused number so ans is the original array
    if ((int) unused.size() == 0) {
        for (int i = 0; i < 2; i++) {
            for (int& x: v) {
                cout << x << " ";
            }
            cout << endl;
        }
        return;
    }

    vector<int> p1(n), p2(n);
   
    // O(n) Get array of lexographical minimum
    p1[0] = v[0];
    auto it = unused.begin();
    for (int i = 1; i < n; i++) {
        if (v[i] != v[i - 1]) {
            p1[i] = v[i];
        } else {
            p1[i] = it->f;
            it++;
        }
    }

    // O(n^2) Get array of lexographical maximum
    p2[0] = v[0];
    for (int i = 1; i < n; i++) {
        if (v[i] != v[i - 1]) {
            p2[i] = v[i];
        } else {
            // find idx
            auto it = unused.lower_bound(p2[i - 1]);
            it--;
            p2[i] = (it)->f;
            unused.erase(p2[i]);
        }
    }

    for (int& x: p1) cout << x << " ";
    cout << endl;
    for (int& x: p2) cout << x << " ";
    cout << endl;
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    srand(chrono::steady_clock::now().time_since_epoch().count());

    int t = 1;
    cin >> t;
    
    while (t--) {
        solve();
    }

    cerr << "time taken : " << (float)clock() / CLOCKS_PER_SEC << " secs" << endl; 
    return 0;
}