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 deb(x) cout << #x << "=" << x << endl
#define deb2(x, y) cout << #x << "=" << x << "," << #y << "=" << y << endl
#define fastio() ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL)
#define gc getchar_unlocked
#define fo(i,n) for( i=0;i<n;i++)
#define pb push_back
#define mp make_pair
#define F first
#define S second
#define all(x) x.begin(), x.end()
#define sortall(x) sort(all(x))
#define tr(it, a) for(auto it = a.begin(); it != a.end(); it++)
#define PI 3.1415926535897932384626
typedef long long ll;
typedef unsigned long long ull;
typedef long double lld;
typedef pair<int, int>	pii;
typedef pair<ll, ll>	pl;
typedef vector<int>		vi;
typedef vector<ll>		vl;
typedef vector<pii>		vpii;
typedef vector<pl>		vpl;
typedef vector<vi>		vvi;

//=================================================================================
//debugger 
struct myComp {
  constexpr bool operator()(double  const& a,
                           double  const& b) const noexcept {
   
    return a>b;
  }
};
#ifndef ONLINE_JUDGE
#define debug(x) cerr << #x <<" "; _print(x); cerr << endl;
#else
#define debug(x)
#endif


void _print(ll t) {cerr << t;}
void _print(int t) {cerr << t;}
void _print(string t) {cerr << t;}
void _print(char t) {cerr << t;}
void _print(lld t) {cerr << t;}
void _print(double t) {cerr << t;}
void _print(ull t) {cerr << t;}

template <class T> void _print (multiset <T,myComp> v);
template <class T, class V> void _print(pair <T, V> p);
template <class T> void _print(vector <T> v);
template <class T> void _print(set <T> v);
template <class T, class V> void _print(map <T, V> v);
template <class T> void _print(multiset <T> v);
template <class T, class V> void _print(pair <T, V> p) {cerr << "{"; _print(p.F); cerr << ","; _print(p.S); cerr << "}";}
template <class T> void _print(vector <T> v) {cerr << "[ "; for (T i : v) {_print(i); cerr << " ";} cerr << "]";}
template <class T> void _print(set <T> v) {cerr << "[ "; for (T i : v) {_print(i); cerr << " ";} cerr << "]";}
template <class T> void _print(multiset <T> v) {cerr << "[ "; for (T i : v) {_print(i); cerr << " ";} cerr << "]";}
template <class T, class V> void _print(map <T, V> v) {cerr << "[ "; for (auto i : v) {_print(i); cerr << " ";} cerr << "]";}
template <class T> void _print (multiset <T,myComp> v) {cerr << "[ "; for (T i : v) {_print(i); cerr << " ";} cerr << "]";}

//=================================================================================================
ll inf =999999999999999;
unsigned int  md = 1e9+7;

//=================================================================================================

void solve ()
{
  //int i,n;
  //cin>>n;
  
 int n,i;
 cin>>n;
 vi a(n),ans1(n),ans2(n);
 set <int>s;
set <int,greater <int> >s2;
 fo (i,n)
 {cin>>a[i];
 s.insert (i+1);
 s2.insert (i+1);
 }
int j=1;
 fo (i,n)
 {
   if ((i==0)||(a[i-1]!=a[i]))
   {ans1[i]=a[i];
   auto it = s.upper_bound (a[i]);
   it --;
   s.erase(it);
   }
   else 
   {auto it = s.begin();
    ans1[i]=(*it);
    s.erase (it);


   }
 }
 fo (i,n)
 {
   if ((i==0)||(a[i]!=a[i-1]))
   {
     ans2[i]=a[i];
     auto it = s2.upper_bound (a[i]);
    
     it-- ;
      debug (*it);
     s2.erase (it );
   }
   else 
   {
     auto it = s2.upper_bound(a[i]);
     ans2[i]= (*it);
      debug (*it);
     s2.erase (it);

   }
 }
 debug (ans1);
 debug (ans2);
 fo (i,n)
 cout << ans1[i]<<" ";
 cout << "\n";
  fo (i,n)
 cout << ans2[i]<<" ";
 cout << "\n";

}
 
//==================================================================================================
int main ()
{

 #ifndef ONLINE_JUDGE
  freopen("input.txt", "r", stdin);
  freopen("output.txt", "w", stdout);
  freopen ("error.txt","w",stderr);
 #endif 
    
fastio();


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

}