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:
//never use vec.size()-1 as this returns unsigned integer
#include <bits/stdc++.h>
#define int long long
#define IOS std::ios::sync_with_stdio(false); cin.tie(NULL);cout.tie(NULL);
#define mod 1000000007
#define lld long double
#define mii map<int, int>
#define mci map<char, int>
#define msi map<string, int>
#define pii pair<int, int>
#define ff first
#define ss second
#define all(x) (x).begin(), (x).end()
#define ll long long
#define fi(i, n) for( int i=0 ; i<n ; i++ )
#define fd(i, n) for( int i=n-1; i>=0 ; i-- )
#define f(i, a, b) for( int i=a ; i<b ; i++ )
#define vi(vec, n) vector<int> vec; vec.resize(n);
#define pb push_back
#define mp make_pair
#define inpi(x)  int x; cin>>x;
#define inpl(x)  long x; cin>>;
#define inpll(x)  long long x; cin>>x;
#define inps(x)  string x; cin >> x;
#define fill(a,b) memset(a, b, sizeof(a))
#define fillv(vec, n) fi(i, n){ inpi(x) vec[i] = x; }
#define disp(vec, n) fi(i, n) cout << vec[i] << " ";
#define v(vec) vector<int> vec;
#define vii vector<pair<int, int>>
#define endl "\n"

using namespace std;
const long long N=100005, INF=2000000000000000000;

v (prime)
bool isprime[N];
void pre()
{
    for(int i=2;i<N;i++)
    {
        if(isprime[i])
        {
            for(int j=i*i;j<N;j+=i)
            isprime[j]=false;
            prime.pb(i);
        }
    }
    return;
}

int32_t main()
{
    IOS;
    //pre();
    int test_case=1;
    cin >> test_case;

    while(test_case--)
    {
        inpi(n)
        vi(q, n)
        fillv(q, n)

        vi(a, n)
        vi(b, n)

        bool vis[n+1];
        fill(vis, false);
        fi(i, n)
            vis[q[i]]=true;

        vector<int> unvis;
        f(i, 1, n+1)
        {
            if(!vis[i])
                unvis.pb(i);
        }

        sort(all(unvis));

        a[0]=b[0]=q[0];

        int i=1, j=0;
        while(i<n)
        {
            if(q[i]==q[i-1])
            {
                a[i]=unvis[j];
                j++;
            }

            else
            {
                a[i]=q[i];
            }
            i++;
        }

        disp(a, n);
        cout << endl;

        i=1;
        while(i<n)
        {
            if(q[i]==q[i-1])
            {
                auto j=lower_bound(all(unvis), q[i-1]);
                j--;
                while( i<n && q[i]==q[i-1] )
                {
                    b[i] = *j;
                    unvis.erase(j);
                    j--;
                    i++;
                }
            }

            else
            {
                b[i]=q[i];
                i++;
            }
        }
        disp(b, n);
        cout << endl;

    }

    return 0;
}