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 <algorithm>
#include <cmath>
#include <vector>
#include <set>
#include <map>
#include <unordered_set>
#include <unordered_map>
#include <queue>
#include <string>
#include <complex>
#include <chrono>
#include <random>
#include <bitset>
#include <iomanip>

using namespace std;

typedef long long ll;
typedef long double ld;
typedef vector<int> vii;
typedef vector<ll> vll;
typedef vector<vll> vvll;
typedef vector<vii> vvii;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef pair<string,int> psi;
typedef vector<pii> vpii;
typedef vector<pll> vpll;
typedef vector<bool> vbb;
typedef vector<string> vss;

const int N = 500001;
const int M = 20;
const int MOD = 1000000007;
const int inf = 1234567890;
const ll INF = 1122334455667788990;

#define fast() ios_base::sync_with_stdio(false),cin.tie(NULL)
#define all(a) a.begin(),a.end()
#define rall(a) a.rbegin(),a.rend()
#define rev(a) reverse(all(a))
#define pb push_back
#define mp make_pair
#define ff first
#define ss second
#define B begin()
#define E end()
#define ub upper_bound
#define lb lower_bound
#define yes cout<<"YES"<<nl
#define no cout<<"NO"<<nl
#define error cout<<-1<<nl
#define space ' '
#define nl '\n'
#define PI acos(-1)
#define sqr(x) ((x)*(x))
#define MAX(x) *max_element(all(x))
#define MIN(x) *min_element(all(x))
#define input freopen("input.txt","r",stdin)
#define output freopen("output.txt","w",stdout)
#define Case(x) cout<<"Case "<<x<<": "
#define dbg(x) cerr<<#x<<" is "<<x<<'\n';
#define fix(x) cout<<fixed<<setprecision(x)

#define coutv(v) for(auto it:v)cout<<it<<' ';cout<<nl;
#define cinv(v) for(auto &it:v)cin>>it;


int main()
{
    fast();
    int tc=1,cs=0;
    cin>>tc;
    while(tc--)
    {
        int n,i;
        cin>>n;
        vii v(n);
        cinv(v);
        vii res(n);
        vbb vis(n+1);
        res[0]=v[0];
        vis[v[0]]=true;
        for(i=1; i<n; i++)
            if(v[i]>v[i-1])
            {
                res[i]=v[i];
                vis[v[i]]=true;
            }
        vii d;
        for(i=1; i<=n; i++)
            if(!vis[i])d.pb(i);
        vii ans=res;
        int j=0;
        for(i=0; i<n; i++)
            if(ans[i]==0)
                ans[i]=d[j++];
        coutv(ans);
        ans=res;
        rev(d);
        int l=0;
        priority_queue<int>tmp;
        for(i=0; i<n; i++)
        {
            if(ans[i]==0)
            {
                ans[i]=tmp.top();
                tmp.pop();
                continue;
            }
            for(j=ans[i]-1; j>l; j--)
                if(!vis[j])
                    tmp.push(j);
            l=ans[i];
        }
        coutv(ans);
    }
}