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 ll         long long
#define fl(i,n)    for(i=0;i<n;i++)
#define fld(i,a,b) for(i=a;i>b;i--)
#define fli(i,a,n) for(i=a;i<n;i++)
#define inp(i,a,n) for(i=0;i<n;i++) {cin>>a[i];}
#define op(i,a,n)  for(i=0;i<n;i++) {cout<<a[i]<<" ";}
#define pb         push_back
#define mp         make_pair
#define mod        1000000007 
#define all(x) 	   x.begin(),x.end()	
#define speedup    ios_base::sync_with_stdio(false),cin.tie(0)
using namespace std;
ll fact(ll n){ if(n==0){return 1;}return (n*fact(n-1))%mod;}
ll power(ll x, ll y){ll out = 1;while (y > 0){if (y & 1) out=(out*x)%mod;y = y>>1;x = (x*x)%mod;}return out%mod;}
ll inverse(ll a){return power(a,mod-2);}
ll gcd(ll a,ll b){if(b==0){return a;}gcd(b,a%b);}
ll a[2000000];
void sieve(){ll i,j;fli(i,2,1000001){if(a[i]==0){for(j=i*i;j<1000001;j+=i){a[j]=1;}}}}
bool compar(const pair<ll,ll> &a, const pair<ll,ll> &b) 
{ 
    if(a.first==b.first)
    return (a.second < b.second);
    else
    return (a.first < b.first);
} 
void cp()
{
   int n;
   cin>>n;
   map<int,int>m1,m3,m2;
   int arr[n],arr1[n],arr2[n];
   for(int i=0;i<n;i++)
   {
       cin>>arr[i];
       m2[arr[i]]++;
       arr1[i]=arr[i];
       arr2[i]=arr[i];
   }
   for(int i=1;i<=n;i++)
   {
       if(m2.find(i)==m2.end())
       {
           m1[i]=1;
           m3[i]=1;
       }
   }
   int i=0;
   while(i<n)
   {
       int j=i+1;
       while(arr[j]==arr[i]&&j<n)
       {
           auto it=m1.upper_bound(0);
           arr1[j]=it->first;
           m1.erase(it->first);
           j++;
       }
       i=j;
   }
   for(int i=0;i<n;i++)
   {
       cout<<arr1[i]<<" ";
   }
   cout<<endl;
   i=0;
   while(i<n)
   {
       int j=i+1;
       while(arr[j]==arr[i]&&j<n)
       {
           auto it=m3.lower_bound(arr[i]);
           it--;
           arr2[j]=it->first;
           m3.erase(it->first);
           j++;
       }
       i=j;
   }
   for(int i=0;i<n;i++)
   {
       cout<<arr2[i]<<" ";
   }
   cout<<endl;
   
}
int main()
{
    speedup;
    int t;
    cin>>t;
    while(t--)
    {
    cp();
    }
}