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>
using namespace std;
#define PRINT(x) if(debug) {cout << #x << ": " << x << endl;}
#define PRINTARR(x) if (debug) {cout<<#x<<": ";for(auto i:x){cout<<i<<' ';}cout<<endl;}
#define ABS(x) ((x>=0)?x:-1*(x))
#define divideCeil(a,b) ((a+b-1) / b)
#define INF 1000000007
#define EPS 1e-9
typedef long long ll;
typedef pair<int,int> ii;
typedef vector<int> vi;
typedef vector<ii> vii;

bool debug;

int main(int argc, char *argv[]){
	ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
    debug = argc > 1;

    int t, n;
    cin >> t;
    while(t--){
        cin >> n;
        vi arr(n+1, 0);
        vi arrMin(n+1, -1), arrMax(n+1, -1);
        vector<bool> mark(n+1, false);
        for (int i=1; i<=n; i++){
            cin >> arr[i];
            if (arr[i] != arr[i-1]){
                arrMin[i] = arrMax[i] = arr[i]; 
                mark[arr[i]] = true;
            }
        }
        vi sisa;
        for (int i = 1; i <= n; i++)
        {
            if(!mark[i]){
                sisa.push_back(i);
            }
        }
        
        PRINTARR(arrMin)
        PRINTARR(arrMax)
        PRINTARR(sisa)

        int idx = 0;
        for (int i = 1; i <= n; i++)
        {
            if (arrMin[i] == -1){
                arrMin[i] = sisa[idx++];
            }
        }
        
        for (int i = 1; i <= n; i++)
        {
            cout << arrMin[i] << ' ';
        } cout << endl;

        priority_queue<int> pq;
        idx = 0;
        for (int i = 1; i <= n; i++)
        {
            if (arrMax[i] == -1){
                while(idx<sisa.size() && sisa[idx] < arr[i]){
                    pq.push(sisa[idx++]);
                }
                arrMax[i] = pq.top(); pq.pop();
            }
        }
        for (int i = 1; i <= n; i++)
        {
            cout << arrMax[i] << ' ';
        } cout << endl;
    }
    
    //cout << n << endl;

    return 0;
}