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 <algorithm>
#include <vector>
#include <queue>
#include <map>
#include <set>
#include <iterator>
#define forn(i,n) for(int i = 0; i < n; i++)
#define pb push_back
#define ul unsigned long
#define ll long long
#define ull unsigned long long

using namespace std;


void solve() {
	int n;
	cin >> n;
	vector<int> q(n), res1(n), res2(n);
	forn(i, n) {
		int qi;
		cin >> qi;
		q[i] = res1[i] = res2[i] = qi;
	}

	set<int> left;
	vector<int> need_replace;
	for (int i = 1; i < n; i++) {
		if (q[i] == q[i - 1]) {
			res1[i] = res2[i] = -1;
		}
	}

	for (int i = 1; i <= (int)q.size(); i++) {
		left.insert(i);
	}

	forn(i, res1.size()) {
		if (res1[i] != -1) {
			left.erase(res1[i]);
		}
		else {
			need_replace.pb(i);
		}
	}

	//Min permutation
	int i = 0;
	for (auto it = left.begin(); it != left.end(); it++) {
		res1[need_replace[i]] = *it;
		i++;
	}
	//Max permutation
	for (int j = 0; j < need_replace.size(); j++) {
		auto it = --left.lower_bound(q[need_replace[j]]);
		res2[need_replace[j]] = *it;
		left.erase(it);
	}
	forn(i, n) cout << res1[i] << " ";
	cout << "\n";
	forn(i, n) cout << res2[i] << " ";
	cout << "\n";
	
}



int main() {
	ios_base::sync_with_stdio(false);
	cin.tie(nullptr); cout.tie(nullptr);
	int t;
	cin >> t;
	while (t--) {
		solve();
	}
	return 0;
}