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 sortall(a) sort(a.begin(),a.end())
#define endl '\n'
#define X first
#define Y second
#define reall(a) reverse(a.begin(),a.end())
#define lb long double
#define forr(i,a) for (int i = 0;i<a;i++)
using namespace std;
const int maxn = 50000+5;
typedef pair<int,int> ii;
typedef pair<long long,long long> LL;
const ll mod =14062008;
const ll base = 31;
int xx[4] = { 0, 1, 0, -1 };
int yy[4] = { 1, 0, -1, 0 };
string step = "URDL";
ll Pow(ll a, ll b){ll res = 1;if(b <=0)return 1;while(b>0){if(b%2){res=(res%mod *a%mod);res%=mod;}a=(a%mod*a%mod)%mod;b/=2;}return res%mod;}
int main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    //freopen("balls.in","r",stdin);
    //freopen("balls.out","w",stdout);
    int t; cin >>  t;
    while(t--){
        int n; cin >> n;
        vector<int>a(n);
        forr(i,n) cin >> a[i];
        vector<int>res1(n),res2(n);
        int max1 = -1;
        set<int>s;
        forr(i,n){
            s.insert(i+1);
        }
        forr(i,n){
            if(a[i] > max1){
                s.erase(a[i]);
                max1 = a[i];
                res1[i] = max1;
            }
            else{
                res1[i] = *s.begin();
                s.erase(*s.begin());
            }
        }
        map<int,int>mp;
        forr(i,n){
            mp[a[i]]++;
        }
        s.clear();
        max1 = -1;
        forr(i,n){
            if(!mp[i+1])
                s.insert(i+1);
        }
        forr(i,n){
             if(a[i] > max1){
                s.erase(a[i]);
                max1 = a[i];
                res2[i] = max1;
            }
            else{
                auto it = s.lower_bound(max1);
                --it;
                res2[i] = *it;
                s.erase(it);
            }
        }
        for(int i : res1) cout << i <<" ";
        cout << endl;
        for(int i : res2) cout << i <<" ";
        cout << endl;

    }
}