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 ll long long
#define pii pair<int,int>
#define mii map<int,int>
#define pli pair<ll,int>
#define pil pair<int,ll>
#define mli map<ll,int>
#define mil map<int,ll>
#define db cout<<"debug"<<endl;
/*
Author:        Rye
Data:        2021-03-25
Email:        Rye10000010000@hotmail.com
*/

const int maxn = 3e5+5;
int a[maxn];
int a1[maxn],a2[maxn];
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int t;
    cin >> t;
    while(t--)
    {
        int n;
        cin >> n;
        map<int,int> occ;
        set<int> st;
        for(int i = 1; i <= n; i++)
        {
            a1[i]=a2[i]=0;
            cin>>a[i];
            if(!occ[a[i]])
            {
                occ[a[i]]=1;
                a1[i] = a2[i] = a[i];
            }
        }
////        pos.push_back(n+1);
//db
        vector<int> pp;
        for(int i = 1; i <= n; i++)
        {
            if(!occ[i]){
                st.insert(i);
                pp.push_back(i);
            }
        }
//        db
//        db
        int now = 0;
        int cnt = 0;
        int pre;
        if(!pp.empty())
        pre = pp[0];
        for(int i = 1; i <= n; i++)
        {
            if(!a1[i])
            {
                a1[i] = pp[now++];
            }
            if(!a2[i])
            {
//                cout<<1<<endl;
                set<int>::iterator pp = st.upper_bound(pre);
                pp--;
                a2[i] = *pp;
                st.erase(pp);
            }else
            {
                pre = a2[i];
            }

        }
        for(int i = 1; i <= n; i++)
            cout<<a1[i]<<" ";
        cout<<endl;
        for(int i = 1; i <= n; i++)
            cout<<a2[i]<<" ";
        cout<<endl;
    }
    return 0;
}