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>

using namespace std;
#define ll                long long
#define pb                push_back
#define mp                make_pair
#define sz(a)             int((a).size())
#define all(c)            (c).begin(),(c).end()
#define	EPS 		      1.0e-8
#define dot(a,b) 	      ((conj(a)*(b)).real())
#define	cross(a,b)	      ((conj(a)*(b)).imag())
#define vec(a,b)	      ((b)-(a))
inline void debugMode() {
	ios_base::sync_with_stdio(false);
    cin.tie(NULL);
}
const int N = 1e5+1;
const int MOD = 1e9+7;
// 1<<m -> 2^m
// Square numbers have odd no. of factors
// (i + j) mod 3 (adjacent diagonals)
// When doing knapsack start from the biggest number and check if depth left is enough
/* Comparator function:
(a < b) increasing
(a > b) decreasing
*/
vector<int> arr,v,high,num;
int n;
int getnum(int x){
	int left = 0,right = num.size();
	while(left + 1 < right){
		int mid = (left + right) / 2;
		if(num[mid]<=x) left = mid;
		else right = mid;
	}
	int nom = num[left];
	num.erase(num.begin() + left);
	return nom;
}
int main()
{
    debugMode();
    int t;cin>>t;
    while(t--){
    	cin>>n;
    	arr.clear();v.clear();high.clear();num.clear();arr.resize(n);v.resize(n);high.resize(n);
    	for(int i=0;i<n;i++) cin>>arr[i];
    	unordered_set<int> s;
    	for(int i=0;i<n;i++){if(!s.count(arr[i])){high[i] = arr[i];v[i] = arr[i];s.insert(arr[i]);}}
    	for(int i=1;i<=n;i++){if(!s.count(i)) num.pb(i);}
    	sort(all(num));
    	for(int i=0,x=1;i<n;i++){
    		if(!v[i]){
    			if(!s.count(x)){v[i] = x;x++;}
    			else {x++;i--;}
    		}
    	}
    	for(int i=0;i<n;i++){cout<<v[i]<<" ";}cout<<endl;
    	for(int i=0;i<n;i++){
    		if(!high[i]){
    			high[i] = getnum(high[i-1]);
    		}
    	}
    	for(int i=0;i<n;i++){cout<<high[i]<<" ";}cout<<endl;
    }
    return 0;
}