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:
/**
 *    author:  Ujjwal Tripathi
 *    created: 27.03.2021 19:19:59
**/

    
#include<bits/stdc++.h>
using namespace std;

#define ar array<ll, 2>
#define ll long long
#define in insert
#define pb push_back
#define pii pair<ll,ll>
#define vt vector
#define fi first
#define se second
#define P_Q(x) priority_queue<x>
#define p_q(x) priority_queue<x, vector<x>, greater<x>>
#define For(i, n) for(ll i = 0; i < n; i++)
#define Rev(i, n) for(ll i = n-1; i >= 0; i--)
#define FOR(i, n) for(ll i = 1; i <= n; i++)
#define REV(i, n) for(ll i = n; i >= 1; i--)
#define Rep(i,a,b) for(ll i = a; i <= b; i++)
#define all(x) (x).begin(),(x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define sz(x) (int) (x).size()

// Maximum priority queue
template <typename T>
using mxpq = priority_queue<T>;

// Minimum priority queue
template <typename T>
using mnpq = priority_queue<T, vt<T>, greater<T>>;


// x | (x + 1) sets lowest unset bit of x
// x & (x - 1) unsets lowest set bit of x

const ll mod = 1e9L + 7;
const ll INF = 1e18L + 5;
const ll mxN = 2e5 + 1;


// remember to read no of test_cases
void solve()
{
   ll n;
   cin>>n;
   vt<ll> v(n,0);
   set<ll> s,s1;
   For(i,n)
   {
    cin>>v[i];
    s.insert(i+1);
    s1.insert(i+1);
   }
    cout<<v[0]<<" ";
   auto it=s.find(v[0]);
   s.erase(it);
   for(ll i=1;i<n;i++)
   {
     if(v[i]==v[i-1])
     {
       it=s.begin();
       cout<<*it<<" ";
       s.erase(it);
     }
     else
     {
       it=s.find(v[i]);
       cout<<v[i]<<" ";
       s.erase(it);
     }
   }
   cout<<endl;
   cout<<v[0]<<" ";
     auto itr=s1.find(v[0]);
   s1.erase(itr);
   for(ll i=1;i<n;i++)
   {
     if(v[i]==v[i-1])
     {
       itr=s1.lower_bound(v[i]);
       if(itr!=s1.begin())
       {
        itr--;
       }
       cout<<*itr<<" ";
       s1.erase(itr);
     }
     else
     {
       itr=s1.find(v[i]);
       cout<<*itr<<" ";
       s1.erase(itr);
     }
   }
   cout<<endl;
  return;
}

int main()
{
  ios::sync_with_stdio(0);  cin.tie(0);
  ll t = 1;
   cin >> t;
  FOR(tt, t) solve();
  return 0;
}