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 <vector>
#include <math.h>
#include<algorithm>
#include <iostream>
#include<set>
#include <string>
#include <math.h>
#include <numeric>
#include<map>
#include<queue>

#define pll pair<ll,ll>
#define pii pair<int,int>
//define ll long long int
#define vi vector<int>
#define vll vector<long long int>
#define ve vector
#define fi first
#define se second
#define rep(i, n) for(int (i)=0;(i)<(n);(i)++)
#define pb push_back
#define len(a) (int)(a).size()
#define all(a) (a).begin(),(a).end()
#define endl  '\n'
#define M_PI 3.14159265358979323846
#define re return
const int INF = 2 * 1e9;
//const ll MOD = 1000000007;
using namespace std;

#include <cassert>

#define int long long

void solve() {
   int n;
   cin>>n;
   vi a(n);
   vi minp(n);
   vi maxp(n);
   vi was(n+1,0);

   rep(i,n) cin>>a[i];
   int used = 1;
   int last = -1;
   rep(i,n) {
       if (last != a[i]) {
           last = a[i];
           was[a[i]] = 1;
           minp[i] = a[i];
       }
       else {
           while (was[used]) used++;
           was[used] = 1;
           minp[i] = used;
       }
   }

    used = 0;
   last = -1;
   vi v(n+1,0);
    rep(i,n) {
        if (last != a[i]) {

            for(int j=last+1;j<a[i];++j)
            {
                v[++used]=j;
            }
            last = a[i];
            maxp[i] = last;
        }
        else {
            maxp[i] = v[used--];
        }
    }

    rep(i,n) cout<<minp[i]<< " ";
    cout<<endl;
    rep(i,n)  cout<<maxp[i]<< " ";
    cout<<endl;
    re;

}


signed main() {
    int t = 1;
    cin >> t;
    while (t--)
        solve();
    re 0;
}