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 <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
#define pb push_back
#define mp make_pair
#define endl "\n"
#define F first
#define S second
#define lb lower_bound
#define ub upper_bound
#define all(x) (x).begin(), (x).end()        //for(auto it=  .begin();it!=  .end();it++)
#define tol(s) transform(s.begin(),s.end(),s.begin(),::tolower);
#define tou(s) transform(s.begin(),s.end(),s.begin(),::toupper);
#define T ll t=0;cin>>t;for(int tttt=0;tttt<t;tttt++)
#define fo(i,a,b) for(i=a;i<b;i++)          
#define of(i,a,b) for(i=a;i>=b;i--)
#define itr(it,a) for(auto it= a.begin();it!= a.end();it++) 
void solve()
{}

int main() {
	 ios_base::sync_with_stdio(false);
     cin.tie(0);cout.tie(0);
     T{
     int n,ans=0,k,x,y,i,j,c=0;
     //string a,b;
     char ch;
     cin>>n;
     set <int> s2,s3;
     int q[n];
     fo(i,0,n)
     {cin>>q[i];}
     int lsd=0;
     
     fo(i,0,n)
     {  fo(k,lsd+1,q[i])s2.insert(k);
         lsd=q[i];
         if( i>0 && q[i]==q[i-1])
         {  j=i;
             while(q[j]==q[i-1])
              {cout<<*s2.begin()<<" ";s2.erase(s2.begin());j++;
              }
          i=--j;
         }
         else
         cout<<q[i]<<" ";
         
      }
      cout<<endl;
     
      lsd=0;
      fo(i,0,n)
     {  fo(k,lsd+1,q[i])s3.insert(k);
        lsd=q[i];
         if( i>0 && q[i]==q[i-1])
         {  j=i;
             while(q[j]==q[i-1])
              {set<int>::iterator it=--s3.end();cout<<*it<<" ";s3.erase(it);j++;
              }
          i=--j;
         }
         else
         cout<<q[i]<<" ";
         
      }
    //  j=0;
    //  fo(i,0,n)
    //  {
    //      if(q[i]!=-1)
    //      cout<<q[i]<<" ";
    //      else 
    //      {
    //          cout<<v[j]<<" ";j++;
    //      }
    //  }
    //  cout<<endl;
    //  reverse(v.begin(),v.end());
    //  fo(i,0,n)
    //  {
    //      if(q[i]!=-1)
    //      {cout<<q[i]<<" ";x=q[i];}
    //      else 
    //      {  j=0;
    //          while(v[j]>x || v[j]==-1)j++;
    //          cout<<v[j]<<" ";v[j]=-1;
    //      }
    //  }
     cout<<endl;
     
    
        
      
}
}