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> pll;
#define F first
#define S second
#define pb push_back
#define mp make_pair
#define all(X) (X).begin(), (X).end()
#define REP(i,x,y) for(ll i = x;i <= y;++i)
#define fastio ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0)
vector<string> vec_splitter(string s) {
    for(char& c: s) c = c == ','?  ' ': c;
    stringstream ss; ss << s;
    vector<string> res;
    for(string z; ss >> z; res.push_back(z))
        ;
    return res;
}
void debug_out(vector<string> args, int idx) { cerr << endl; }
template <typename Head, typename... Tail>
void debug_out(vector<string> args, int idx, Head H, Tail... T) {
    if(idx > 0) cerr << ", ";
    stringstream ss; ss << H;
    cerr << args[idx] << " = " << ss.str();
    debug_out(args, idx + 1, T...);
}
#define debug(...) debug_out(vec_splitter(#__VA_ARGS__), 0, __VA_ARGS__)
void localTest() {
    #ifndef ONLINE_JUDGE
        freopen("inp", "r", stdin);
        freopen("out", "w", stdout);
    #endif      
}
void solver() {
    ll n;
    cin >> n;
    vector<ll> A(n);
    set<ll> st, tmp;
    for(ll i = 1; i <= n; ++i) {
        st.insert(i);
    }
    for(auto &x : A) {
        cin >> x;
        if(st.find(x) != st.end()) {
            st.erase(x);
        }
    }   
    tmp = st;
    cout << A[0] << " ";
    for(ll i = 1; i < n; ++i) {
        if(A[i] == A[i - 1]) {
            cout << *st.begin() << " ";
            st.erase(*st.begin());
        } else {
            cout << A[i] << " ";
        }
    }
    cout << "\n";
    st = tmp;
    cout << A[0] << " ";
    for(ll i = 1; i < n; ++i) {
        if(A[i] == A[i - 1]) {
            auto lo = st.lower_bound(A[i]);
            --lo;
            cout << *lo << " ";
            st.erase(*lo);
        } else {
            cout << A[i] << " ";
        }
    }
    cout << "\n";

}
int main() {

    localTest();
    fastio;
    ll t;
    cin >> t;
    while(t--) {
        solver();
    }

    return 0;
}