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:
/*
    Author: mr_cruise/krishrustagi
    
    Rules:
    1. Work for satisfaction and not for hunger.
    2. Read the question carefully.
    3. Check for the corner cases.
*/

#include <bits/stdc++.h> 
using namespace std;
 
typedef long long ll; 
#define fori(i, n) for (ll(i) = 0; (i) < (n); (i)++) 
#define foria(i, a, b) for (ll(i) = (a); (i) <= (b); (i)++) 
#define foriar(i, a, b) for (ll(i) = (a); (i) >= (b); (i)--)  
#define ff first 
#define ss second 
#define pb push_back 
#define endl '\n'

const ll mod = 1e9 + 7;
const ll imx = INT_MAX;
const ll imn = INT_MIN;
 
int static dp[1001][1001];

void solve()
{
    ll n;
    cin >> n;

    ll a[n];
    set<ll> s, s1;
    fori(i, n)
    {
        cin >> a[i];
        s.insert(i+1);
        s1.insert(i+1);
    }

    fori(i, n)
    {
        if(s.find(a[i]) != s.end())
        {
            s.erase(a[i]);
            s1.erase(a[i]);
        }
    }

    ll ans1[n];
    ans1[0] = a[0];
    foria(i, 1, n-1)
    {
        if(a[i] != a[i-1])
        {
            ans1[i] = a[i];
        }
        else
        {
            ans1[i] = *s.begin();
            s.erase(s.begin());
        }
    }

    fori(i, n)
    {
        cout << ans1[i] << " ";
    }
    cout << endl;

    s1.insert(n+1);
    ll ans2[n];
    ans2[0] = a[0];
    foria(i, 1, n-1)
    {
        if(a[i] != a[i-1])
        {
            ans2[i] = a[i];
        }
        else
        {
            auto x = s1.lower_bound(a[i]);
            // cerr << *x << endl;
            x--;
            // cerr << *x << endl;
            ans2[i] = *x;
            s1.erase(*x);
        }
    }

    fori(i, n)
    {
        cout << ans2[i] << " ";
    }
    cout << endl;
}

int main()
{
    ios_base::sync_with_stdio(false); 
    cin.tie(NULL); 
    cout.tie(NULL);
 
    int t = 1;
    cin >> t;
    fori(_, t)
    {
        // cout << "Case #" << _ + 1 << ": ";
        solve();
    }
    return 0;
}