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>
#define FOR(i,x,y) for ( int i=(x);i<=(y);i++)
#define forr(i,x,y) for ( int i=(x);i<(y);i++)
#define dow(i,x,y) for ( int i=(x);i>=(y);i--)
#define Debug(v) for (auto i:v) cout<<i<<" ";puts("")
#define mp make_pair
#define fi first
#define se second
#define pb push_back
#define ep emplace_back
#define siz(x) ((int)(x).size())
#define all(x) (x).begin(),(x).end()
#define fil(a,b) memset((a),(b),sizeof(a))
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pa;
typedef pair<ll,ll> PA;
typedef vector<int> poly;
//const int N = 2e5+10, M = 5e6+10;

void solve(){
int n;
cin>>n;
vector<int> a(n,0);
vector<int> b(n+1,0);
vector<int> c(n,0);
set<int> s1,s2;
forr(i,0,n){
	cin>>a[i];
	s1.insert(a[i]);
	}
	forr(i,1,n+1){
	if (s1.find(i)==s1.end()){s1.insert(i);}
	else s1.erase(i);
	}
	
	b[0]=a[0];c[0]=b[0];
	
	
	int j=0;
cout<<a[0]<<" ";
forr(i,1,n){
	if(a[i]>a[i-1]){b[i]=a[i];}
	else  {b[i]=*s1.begin();s1.erase(*s1.begin());	}
	cout<<b[i]<<" ";}
	
cout<<endl<<a[0]<<" ";
		forr(i,1,a[0]){
			s2.insert(i);}
			
forr(i,1,n){
	if(a[i]>a[i-1]){b[i]=a[i];
		forr(j,a[i-1]+1,a[i]){
	s2.insert(j);
	}}
	else  {b[i]=*s2.rbegin();s2.erase(b[i]);	}
	cout<<b[i]<<" ";
	}
	cout<<endl;
	}

	
int main(){
	int t=1;
	cin>>t;
	while(t--){
		solve();}
	
	return 0;
	}