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 <map>
#include <cmath>
#include <string>
#include <set>
#include <unordered_set>
#include <algorithm>
#include <unordered_map>
using namespace std;

#define lli long long int
#define ll long long
#define sp ' '
#define nl '\n'
#define push push_back
#define pop pop_back
#define make make_pair
#define forn(i,j,k) for (ll i = j; i <= k; ++i)
#define rev_for(i,j,k) for(ll i = j;i >= k; i--)
#define for_(i,n) for(ll i=0;i < n; i++)
#define r_for(i,n) for(ll i=n-1;i >= 0; i--)
#define inc_for(i,j,k,l) for (ll i = j; i <= k; i+=l)
#define dcr_for(i,j,k,l) for (ll i = j; i >= k; i-=l)
//TODO:check the constraint lli or int
//becasuse i always.....
void test_cases(){
	int n;
	cin>>n;
	set<int> c;
	vector<int> min(n),in,max(n);
	forn(i,1,n){
		c.insert(i);
		min[i-1]=0;
	}
	vector<int> a(n);
	int p=0;
	for_(i,n){
		cin>>a[i];
		if(a[i]!=p){
			p=a[i];
			c.erase(p);
			min[i]=p;
			max[i]=p;
		}else{
			in.push(i);
		}
	}
	int m=0;
	for (std::set<int>::iterator i = c.begin(); i != c.end(); ++i)
	{
		min[in[m]]=*i;
		m++;
	}
	m=0;
	for_(i,in.size()){
		std::set<int>::iterator result =c.upper_bound(a[in[i]]);
		--result;
    	a[in[i]]=*result;
    	c.erase(*result);
	}
	for_(i,n){
		cout<<min[i]<<sp;
	}
	cout<<nl; 
	for_(i,n){
		cout<<a[i]<<sp;
	}
}
int main(int argc, char const *argv[])
{
	int n=1;
	cin>>n;
	forn(i,1,n){
		// cout<<"Case #"<<i<<": ";
		test_cases();
		cout<<nl;
	}
	return 0;
}