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 <bits/stdc++.h>
//#include <ext/pb_ds/tree_policy.hpp>
//#include <ext/pb_ds/assoc_container.hpp>

using namespace std;
//using namespace __gnu_pbds;

#define debug(x) cerr << #x << " :: " << x << '\n';

const int N = 3e5 + 10;
const int MOD = 1e9 + 7;
const long long INF = 5e18;
const double PI = 2 * acos(0.0);
//template<typename T> using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int t;
    cin >> t;
    while (t--) {
        int n;
        cin >> n;
        vector<int> a(n);
        for (auto &i : a) {
            cin >> i;
        }
        set<int> s;
        for (int i = 1; i <= n; i++) {
            s.insert(i);
        }
        int mx = 0;
        set<int> have;
        vector<int> minAns, maxAns;
        for (int i = 0; i < n; i++) {
            if (a[i] > mx) {
                mx = a[i];
                while (s.size() > 0 && *s.begin() <= mx) {
                    have.insert(*s.begin());
                    s.erase(*s.begin());
                }
                minAns.push_back(mx);
                have.erase(mx);
            } else {
                minAns.push_back(*have.begin());
                have.erase(*have.begin());
            }
        }
        mx = 0;
        s.clear();
        have.clear();
        for (int i = 1; i <= n; i++) {
            s.insert(i);
        }
        for (int i = 0; i < n; i++) {
            if (a[i] > mx) {
                mx = a[i];
                while (s.size() > 0 && *s.begin() <= mx) {
                    have.insert(*s.begin());
                    s.erase(*s.begin());
                }
                maxAns.push_back(mx);
                have.erase(mx);
            } else {
                maxAns.push_back(*have.rbegin());
                have.erase(*have.rbegin());
            }
        }
        for (auto i : minAns) {
            cout << i << " ";
        }
        cout << '\n';
        for (auto i : maxAns) {
            cout << i << " ";
        }
        cout << '\n';
    }
    return 0;
}