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>
#define  na
#define ll long long 
#define  ff first
#define ss second
#define vi vector<int>
#define vl vector<ll>
#define vll vector<pair<ll,ll> >
#define vvi vector<vector<int> >
#define vii vector<pair<int,int> >
#define pb push_back
#define inf 1000000007
#define ninf -1000000007
#define mod 998244353
#define f(i,a,b) for(i=a;i<b;i++)
#define re0 return 0
using namespace std;
vector<set<int> > v(10001);
int vis[10001];
int dfs(int par)
{
	vis[par]=1;
	
	for(auto node:v[par])
	{
		if(vis[node]==0)
		{
			dfs(node);
		}
	}
}
bool cmp(const string& a,const string& b)
{
	 return (a+b<b+a);
}
ll gcdextended(ll a, ll b, ll *x, ll *y)
{
	 if(a==0)
	 {
	 	*x=0,*y=1;
	 	return b;
	 }
	 ll x1,y1;
	 
	 ll gcd=gcdextended(b%a,a,&x1,&y1);
	 
	 *x=y1-(b/a)*x1;
	 *y=x1;
	 return gcd;
}
ll modinverse(ll a,ll m)
{
	ll x,y;
	
	ll g=gcdextended(a, m, &x, &y);
	
	return ((x%m+m)%m);
}

ll nCr(ll n, ll r, ll p) 
{ 
    // Optimization for the cases when r is large 
    if (r > n - r) 
        r = n - r; 
  
    // The array C is going to store last row of 
    // pascal triangle at the end. And last entry 
    // of last row is nCr 
    ll C[r + 1]; 
    for(ll i=0;i<=r;i++)C[i]=0;
  
    C[0] = 1; // Top row of Pascal Triangle 
  
    // One by constructs remaining rows of Pascal 
    // Triangle from top to bottom 
    for (ll i = 1; i <= n; i++) { 
  
        // Fill entries of current row using previous 
        // row values 
        for (ll j = min(i, r); j > 0; j--) 
  
            // nCj = (n-1)Cj + (n-1)C(j-1); 
            C[j] = (C[j] + C[j - 1]) % p; 
    } 
    return C[r]%inf; 
} 
ll exp(ll n,ll j)
{
	ll i=n,res=1;
	
	while(j>0)
	{
		if(j&1)res*=i;
		res%=inf;
		 i=i*i;
		 i%=inf;
		j>>=1;
	}

	return res%inf;
}

int main()
{
	#ifdef ha
	freopen("input.txt","r",stdin);
	freopen("output.txt","w",stdout);
	#endif
 
  int T=1;
  
    scanf("%d",&T);
    
   
    while(T--)
    {
    	int n,i,j;
    	
    	cin>>n;
    	set<int>s,t;
    	int a[n];f(i,0,n)cin>>a[i],s.insert(i+1),t.insert(i+1);
    	
    	f(i,0,n)
    	{
    		auto it=s.find(a[i]);
    		
    		if(it!=s.end())
    		{
    			cout<<a[i]<<' ';s.erase(it);
			}
			else
			{
				cout<<*s.begin()<<' ';
				s.erase(s.begin());
			}
		}
		cout<<'\n';
		f(i,0,n)
		{
			auto it=t.upper_bound(a[i]);
			it--;
			cout<<*it<<' ';
			t.erase(it);
		}
		cout<<'\n';
	}
}