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 inf          1e15
#define mod          1000000007
#define N            100001
#define ll           long long int
#define vint         vector<int>
#define vll          vector<ll>
#define vstr         vector<string>
#define vvint        vector<vector<int> >
#define vvll         vector<vector<ll> >
#define vint_pair    vector<pair<int,int> >
#define input(arr)   for(auto &x:arr) cin>>x;
#define all(x)       (x).begin(), (x).end()
#define rall(x)      (x).rbegin(), (x).rend()

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int t=1;
    cin>>t;
    while(t--){
        int n;
        cin>>n;
        set<int>num;
        vint arr(n);
        for(int i=0;i<n;++i){
            cin>>arr[i];
            num.insert(arr[i]);
        }
        int vis[n+1];
        memset(vis, 0, sizeof(vis));
        int l = 1;
        for(int i=0;i<n;++i){
            if(vis[arr[i]]){
                cout<<l<<" ";
                vis[l] = 1;
            }
            else{
                cout<<arr[i]<<" ";
                vis[arr[i]] = 1;
            }
            while(vis[l]){
                ++l;
            }
        }
        cout<<'\n';
        set<int>st;
        memset(vis, 0, sizeof(vis));
        for(int i=1;i<=n;++i){
            if(num.find(i)==num.end()){
                st.insert(i);
            }
        }
        for(int i=0;i<n;++i){
            if(!vis[arr[i]]){
                cout<<arr[i]<<" ";
                vis[arr[i]] = 1;
            }
            else{
                auto it = st.lower_bound(arr[i]);
                --it;
                int tmp = *it;
                cout<<tmp<<" ";
                vis[tmp] = 1;
                st.erase(tmp);
            }
        }
        cout<<'\n';
    }
}