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;
using ll=long long;
#define fi first
#define se second
#define rep(i,n) for(int i=0;i<n;i++)
#define all(v) v.begin(),v.end()
#define pb push_back
template<class T,class U>
inline bool chmax(T &a,U b){
    if(a<b){
        a=b;
        return true;
    }
    return false;
}
template<class T,class U>
inline bool chmin(T &a,U b){
    if(a>b){
        a=b;
        return true;
    }
    return false;
}
constexpr int INF=1000000000;
constexpr ll llINF=3000000000000000000;
constexpr int mod=998244353;
constexpr double eps=1e-8;
const double pi=acos(-1);
int dx[]={0,1,0,-1},dy[]={1,0,-1,0};
int Random(int mi,int ma){
    random_device rnd;
    mt19937 mt(rnd());//32bit
    //[mi,ma]
    uniform_int_distribution<int>engine(mi,ma);
    return engine(mt);
}
/*
vector<vector<ll>>C,sC;
void init_comb(int n,int m){
    C.resize(n+1,vector<ll>(m+1,0));
    sC.resize(n+1,vector<ll>(m+1,0));
    C[0][0]=1;
    for(int i=1;i<=n;i++){
        C[i][0]=1;
        for(int j=1;j<=m;j++){
            C[i][j]=(C[i-1][j-1]+C[i-1][j])%mod;
        }
    }
    rep(i,n+1){
        rep(j,m){
            sC[i][j+1]=(sC[i][j]+C[i][j])%mod;
        }
    }
}*/
ll gcd(ll a,ll b){
    while(a%b){
        a%=b;
        swap(a,b);
    }
    return b;
}
ll lcm(ll a,ll b){
    return a/gcd(a,b)*b;
}
bool isprime(int a){
    if(a==1)return false;
    for(int i=2;i*i<=a;i++){
        if(a%i==0)return false;
    }
    return true;
}
vector<int>primes;
void init_prime(int n){
	primes.push_back(2);
    for(int i=3;i<=n;i+=2){
        bool f=true;
        for(int j:primes){
            if(j*j>i)break;
            if(i%j==0){
                f=false;
                break;
            }
        }
        if(f)primes.push_back(i);
    }
}
ll modpow(ll a,ll b){
    ll res=1;
    while(b){
        if(b&1){
            res*=a;
            res%=mod;
        }
        a*=a;
        a%=mod;
        b>>=1;
    }
    return res;
}
vector<ll>inv,fact,factinv;
void init_fact(int n){
    inv.resize(n+1);
    fact.resize(n+1);
    factinv.resize(n+1);
    inv[0]=0;
    inv[1]=1;
    fact[0]=1;
    factinv[0]=1;
    for(ll i=1;i<=n;i++){
        if(i>=2)inv[i]=mod-((mod/i)*inv[mod%i]%mod);
        fact[i]=(fact[i-1]*i)%mod;
        factinv[i]=factinv[i-1]*inv[i]%mod;
    }
}
ll _inv(ll a,ll m=mod){
    //gcd(a,m) must be 1
    ll b=m,u=1,v=0;
    while(b){
        ll t=a/b;
        a-=t*b;swap(a,b);
        u-=t*v;swap(u,v);
    }
    u%=m;
    if(u<0)u+=m;
    return u;
}
ll comb(int a,int b){
    if(a<b)return 0;
    if(a<0)return 0;
    return fact[a]*factinv[a-b]%mod*factinv[b]%mod;
}
ll multicomb(int a,int b){
    return comb(a+b-1,b);
}

void solve(){
    int n;
    cin>>n;
    vector<int>a(n),mi(n),ma(n);
    for(int &x:a)cin>>x;
    set<int>st;
    for(int i=1;i<=n;i++)st.insert(i);
    rep(i,n){
        if(i==0||a[i-1]<a[i]){
            mi[i]=a[i];
            st.erase(a[i]);
        }
        else{
            mi[i]=*st.begin();
            st.erase(st.begin());
        }
    }
    for(int i=1;i<=n;i++)st.insert(i);
    rep(i,n){
        if(i==0||a[i-1]<a[i]){
            ma[i]=a[i];
            st.erase(a[i]);
        }
        else{
            auto it=st.lower_bound(a[i]);
            it--;
            ma[i]=*it;
            st.erase(it);
        }
    }
    for(int i:mi)cout<<i<<" ";
    cout<<endl;
    for(int i:ma)cout<<i<<" ";
    cout<<endl;
}
int main(){
    cin.tie(0);ios::sync_with_stdio(false);
    int t;cin>>t;while(t--)
    solve();
    return 0;
}