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:
// RAKSHIT KADAM
 
//  -- Declare array globally...
 #include <bits/stdc++.h>
// #include <ext/pb_ds/assoc_container.hpp>
// #include <ext/pb_ds/tree_policy.hpp>
// using namespace __gnu_pbds;
// #define ordered_set tree< int ,  null_type ,  less<int> ,  rb_tree_tag ,  tree_order_statistics_node_update>
// find_by_order returns iterator to  element present at that index...
// order_of_key returns no. of elements having value stricktly less than that given in the order_of_key();
#warning !!!check the size of arrayss!!!
 
// NUMERIC DEFINITIONS :
  
    #define inf 1000000003
    #define MOD  1000000007
    #define newMOD  998244353
    #define MAX 200006
    #define pi 3.1415926535897932384626433832795
 #define ld long doubles
  #define ll long long
  #define int long long
 #define INF 1000000000000000008
 #define Need_for_speed(activated) ios_base :: sync_with_stdio(0);cin.tie(NULL);cout.tie(NULL)
 #define satisfy ll t; cin>> t; while(t--)
 #define mp make_pair
 #define pb push_back
 #define endl "\n"
 #define x first
 #define y second
 #define Endl endl
 #define pii pair<ll,ll>
 #define input freopen("input.txt", "r", stdin),freopen("output.txt", "w", stdout)
 #define time        cerr << "\nTime elapsed: " << 1000 * clock() / CLOCKS_PER_SEC << "ms\n";
using namespace std;
ll gcd(ll a, ll b){if (b == 0)return a;return gcd(b, a % b);}
int power(int x, unsigned int y, unsigned int m){ if (y == 0) return 1;int p = power(x, y/2, m) % m;  p = (p * p) % m;return (y%2 == 0)? p : (x * p) % m;}
ll modInverse(int a, int m){{ return power(a, m-2, m);}}
const int maxn = 500004;
// ll fact[300007] = {0};
// void facto() {fact[0] = 1; fact[1] = 1; for(int i = 2; i < 300005; i++){fact[i] = (fact[i-1]*i)%MOD;}}
// ll ncr(ll n, ll r) {
//     if(r>n)return 0;
//     ll res = 1; res = fact[n]; res = (res*(fpow(fact[r], MOD-2)))%MOD; res = (res*(fpow(fact[n-r], MOD-2)))%MOD; return res;
// }
 

void solve()
{
    ll n;cin>>n;
    ll a[n];  set<ll>s,b;
    for(ll i=0;i<n;i++)
    {
        cin>>a[i];
        s.insert(i+1);
    }
  
    ll aa[n],bb[n];
    memset(aa,0,sizeof aa);
    memset(bb,0,sizeof bb);
    ll x =a[0];
    aa[0]=bb[0]=x;
    s.erase(a[0]);
    for(ll i=1;i<n;i++)
    {
        if(a[i]!=x)
        { aa[i]=bb[i]=a[i];x=a[i];s.erase(s.find(a[i]));}
    }
    for(auto it:s)b.insert(it);
    for(ll i=0;i<n;i++)
    {
        if(bb[i])
        {}
        else
        {
            auto it  = s.lower_bound(a[i]);
            it--;
            bb[i]=*it;
            s.erase(it);
        }
    }
    for(ll i=0;i<n;i++)
        if(!aa[i])
        {
            aa[i]=*b.begin();
            b.erase(b.begin());
        }
    for(ll i=0;i<n;i++)
        cout<<aa[i]<<" ";
    cout<<endl;
    for(ll i=0;i<n;i++)cout<<bb[i]<<" ";
    cout<<endl;
}
 
signed main()
{
    
   Need_for_speed(activated);
satisfy
    { solve();
   }


//    solve();
    return 0;
}