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 DEBUG(x) cout<<"->"<<" x"<<":"<<x<<"\n";
#define FIO ios_base::sync_with_stdio(false);cin.tie(NULL);
using namespace std;
long long int mod=1e9+7;

int main()
{
    FIO;
    int t;
    cin>>t;
    for(int z=0;z<t;z++)
    {
        int n;
        cin>>n;
        int arr[n];
        int ansMin[n];
        memset(ansMin,-1,sizeof(ansMin));
        for(int i=0;i<n;i++)
        {
            cin>>arr[i];
        }
        int ansMax[n];
        memset(ansMax,-1,sizeof(ansMax));
        set<int>taken;
        ansMax[0]=arr[0];
        ansMin[0]=arr[0];
        taken.insert(arr[0]);
        for(int i=1;i<n;i++)
        {
            if(arr[i]!=arr[i-1])
            {
                ansMax[i]=arr[i];
                ansMin[i]=arr[i];
                taken.insert(arr[i]);
            }
        }

        int curr=1;
        for(int i=0;i<n;i++)
        {
            if(ansMin[i]==-1)
            {
                while(taken.find(curr)!=taken.end())
                {
                    curr++;
                }
                ansMin[i]=curr;
                curr++;
            }
        }

        for(int i=0;i<n;i++)
        {
            cout<<ansMin[i]<<" ";
        }
        cout<<"\n";

        set<int>weHave;
        for(int i=1;i<=n;i++)
        {
            if(taken.find(i)==taken.end())
                weHave.insert(i);
        }
        auto it=taken.begin();
        int prev=n;
        for(int i=0;i<n;i++)
        {
            if(ansMax[i]==-1)
            {
                it=taken.begin();
                if(it!=taken.end())
                {
                auto it2=weHave.upper_bound(prev);
                it2--;
                ansMax[i]=*it2;
                weHave.erase(it2);
                }
                else
                {
                    auto it2=weHave.end();
                    it2--;
                    ansMax[i]=*it2;
                    weHave.erase(it2);
                }
            }
            else
            {
                taken.erase(ansMax[i]);
                prev=ansMax[i];
            }
        }

        for(int i=0;i<n;i++)
        {
            cout<<ansMax[i]<<" ";
        }
        cout<<"\n";
    }
    return 0;
}