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 pair<ll,ll> pi;
typedef vector<ll> vi;
typedef vector<pi> vpi;
typedef long double ld;
const int inf = 1e9 + 7;
#define mod 1000000007
const int maxn = 1e6 + 3;
#define pb push_back
#define mp make_pair
#define ff first
#define ss second
#define rep(i, a, b) for(int i = a; i < b; i++)
#define all(v) v.begin(),v.end()
#define rall(v) v.rbegin(),v.rend()
#define lb lower_bound
#define ub upper_bound
#define fact(n) rep(i, 1, n+1)ft.pb((ft[i-1]*i)%mod);
#define mod_in(a) exp(a, mod-2)
#define ncr(n, r) ((ft[n]*mod_in((ft[r]*ft[(n)-(r)])%mod))%mod)
#define deb(x) cout << #x << " " << x << endl;
#define flash ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define endl '\n'

/*#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
template<typename T>
using ordered_set = tree<T,null_type,less<T>,rb_tree_tag,tree_order_statistics_node_update>;
template<typename T>
using ordered_multiset = tree<T,null_type,less_equal<T>,rb_tree_tag,tree_order_statistics_node_update>;*/

ll prime[4000005];
ll exp(ll a,ll b);
void sieve(ll n);
ll gcd(ll a ,ll b);
ll countdigit(ll n);
vector<ll>ft;
ll dp[100005];

int main()
{   
    flash;
    ll t=1;
   // sieve(4000003);
    //ft.pb(1);
    //fact(10000001);
    cin >> t;
    while(t--) {
        ll n;
        cin >> n;
        ll a[n];
        rep(i,0,n) cin >> a[i];
        unordered_map<ll,ll>seedha;
        unordered_map<ll,ll>ulta;
        vector<ll>ans1,ans2(n);
        set<ll>se;
        for(ll i = 0; i < n;i++){
            se.insert(i);
            ulta[a[i]]++;
        }
        stack<ll>st;
        queue<ll>st2;
        for(ll i = n; i >= 1; i--) {
            if(ulta.find(i) == ulta.end()){st.push(i);}
        }
        ll xx = a[n-1];
        ll yy = a[0];
        ll h = yy-1;
        while(h > 0){
            if(ulta.find(h) == ulta.end()){st2.push(h);ulta[h]++;}
            h--;
        }
        h = xx-1;
        while(h > 0){
            if(ulta.find(h) == ulta.end()){st2.push(h);}
            h--;
        }
        ans1.pb(a[0]); 
        for(ll i = 1; i < n; i++) {
            if(a[i] != a[i-1]){
                ans1.pb(a[i]);
            }
            else {
                if(st.size() > 0){
                    ll x = st.top();
                    ans1.pb(x);
                    st.pop();
                }
            }
        }
        for(ll i=0;i<n;i++)
	    {
	        ll j = i+1;
	        ans2[i]  = a[i];
	        
	        while(j<n && a[j]==a[i])
	        {
	            auto it = se.lower_bound(a[i]);
    	        it--;
    	        ll s = *it;
	            ans2[j] = s;
	            se.erase(s);
	            j++;
	        }
	        se.erase(a[i]);
	        i = j-1;
	    }
        for(auto it : ans1){
            cout << it << " ";
        }
        cout << endl;
       // reverse(all(ans2));
        for(auto it : ans2){
            cout << it << " ";
        }
        cout << endl;
        
        
    }
}

//
//
//
//
//
// IMPORTANT FUNCTIONS
ll exp(ll a, ll b) {
	int res=1;
	while(b) {
		if(b&1)
			res=(res*a)%mod;
		a=(a*a)%mod;
		b>>=1;
	}
	return res%mod;
}
ll gcd(ll a,ll b){
    if(a==0)return b;
    return gcd(b%a,a);
}
void sieve(ll n) {
    for(ll i=0;i<=n;i++) {
        prime[i]=1;
    }
    prime[0]=0;prime[1]=0;
    for(ll i=2;i<=n;i++) {
        if(prime[i]==1) {
            for(ll j=2;i*j<=n;j++) {
                prime[i*j]=0;
            }
        }
    }
}
ll countdigit(ll  n)
{
    return floor(log10(n) + 1);
}