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;
 
typedef long long ll;
typedef long double ld;
typedef complex<ld> cd;
 
typedef pair<int, int> pi;
typedef pair<ll,ll> pl;
typedef pair<ld,ld> pd;
 
typedef vector<int> vi;
typedef vector<ld> vd;
typedef vector<ll> vl;
typedef vector<pi> vpi;
typedef vector<pl> vpl;
typedef vector<cd> vcd;
#define endl "\n"
#define int long long
#define FOR(i, a, b) for (int i=a; i<(b); i++)
#define F0R(i, a) for (int i=0; i<(a); i++)
#define FORd(i,a,b) for (int i = (b)-1; i >= a; i--)
#define F0Rd(i,a) for (int i = (a)-1; i >= 0; i--)
#define trav(a,x) for (auto& a : x)
#define uid(a, b) uniform_int_distribution<int>(a, b)(rng)
 
#define sz(x) (int)(x).size()
#define mp make_pair
#define pb push_back
#define f first
#define s second
#define lb lower_bound
#define ub upper_bound
#define all(x) x.begin(), x.end()
#define ins insert
 
template<class T> bool ckmin(T& a, const T& b) { return b < a ? a = b, 1 : 0; }
template<class T> bool ckmax(T& a, const T& b) { return a < b ? a = b, 1 : 0; }
 
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
 
const int MOD = 1000000007;
const char nl = '\n';
const int MX = 100001; //check the limits, dummy
int powi(int x,int n)
{
    int res=1;
    while(n)
    {
        if(n&1) res=(res*x)%MOD;
        x=(x*x)%MOD;
        n>>=1;
    }
    return res;
}
bool isPowerOfTwo(int n) 
{ 
    if (n == 0) 
        return 0; 
    while (n != 1) 
    { 
        if (n%2 != 0) 
            return 0; 
        n = n/2; 
    } 
    return 1; 
} 
int gcd(int a , int b)
{
   if(b==0) return a;
   a%=b;
   return gcd(b,a);
}
void solve(){
  int n;cin>>n;
  int q[n];
  map<int,int> m;
  set<int> s;
  
  FOR(i,0,n)
  {cin>>q[i];
    m[q[i]]++;
    
  }
  vector<int> start;

  FOR(i,1,n+1)
  {
    if(m[i] == 0)
    {
      start.pb(i);
      s.ins(-i);
    }
  }

  vector<int> minans;
  vector<int> maxans;
  minans.pb(q[0]);
  int j =0;

  

  FOR(i,1,n)
  {
    if(q[i] == q[i-1])
    {
       minans.pb(start[j]);
       j++;
    }
    else
    {
      minans.pb(q[i]);
    }
  }
  maxans.pb(q[0]);
  int last =q[0];

  FOR(i,1,n)
  {
    if(q[i] != last)
    {
       maxans.pb(q[i]);
       last = q[i];
    }
    else
    {
      int  b = -(*s.lb(-q[i]));
      maxans.pb(b);
      s.erase(s.lb(-q[i]));

    }
  }
  

  trav(x,minans) cout<<x<<" ";

  cout<<endl;

  trav(x,maxans) cout<<x<<" ";

  cout<<endl;




   

 



  
}

signed main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);
    #ifndef ONLINE_JUDGE
    freopen("input1.txt", "r", stdin);
    freopen("output2.txt", "w", stdout);
    #endif
    
    int t=1;
    cin>>t;
    while(t--) solve();
 
 
    return 0;
}