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:
// set vim: fdm=marker sw=4 ts=4 et

#define _CRT_DISABLE_PERFCRIT_LOCKS
#include <bits/stdc++.h>

using namespace std;

typedef int_fast64_t ll;
// const ll modulo = 1000000007;

#ifndef countof
#define countof(x) (sizeof(x) / sizeof(*(x)))
#endif

// --------------------------------------------------------------------------------

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    size_t T;
    cin >> T;
    for (size_t t = 0; t < T; ++t) {
        int n;
        cin >> n;

        vector<ll> q;
        q.reserve(n);
        copy_n(istream_iterator<ll>(cin), n, back_inserter(q));

        vector<ll> p_min, p_max;
        p_min.reserve(n);
        p_max.reserve(n);

        set<ll> min_avail;
        set<ll, std::greater<ll>> max_avail;
        for (ll i = 1; i <= n; ++i) {
            min_avail.insert(i);
            max_avail.insert(i);
        }

        ll i = 0;
        while (i < n) {
            ll j = i;
            while (j < n && q[j] == q[i]) {
                ++j;
            }

            p_min.push_back(q[i]);
            min_avail.erase(q[i]);
            auto it_min = min_avail.begin();

            auto it_max = max_avail.find(q[i]);
            p_max.push_back(*it_max);
            it_max = max_avail.erase(it_max);

            for (ll k = 0; k < j - i - 1; ++k) {
                p_min.push_back(*it_min);
                it_min = min_avail.erase(it_min);

                p_max.push_back(*it_max);
                it_max = max_avail.erase(it_max);
            }

            i = j;
        }

        copy(begin(p_min), end(p_min), ostream_iterator<ll>(cout, " "));
        cout << endl;
        copy(begin(p_max), end(p_max), ostream_iterator<ll>(cout, " "));
        cout << endl;
    }
    return 0;
}

// --------------------------------------------------------------------------------