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:
//#pragma GCC optimize("Ofast")

#include <bits/stdc++.h>

using namespace std;

void debug_out() { cerr << endl; }

template<typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
    cerr << "[" << H << "]";
    debug_out(T...);
}

#ifdef dddxxz
#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
#else
#define debug(...) 42
#endif

#define SZ(s) ((int)s.size())
#define all(x) (x).begin(), (x).end()
#define revall(x) (x).rbegin(), (x).rend()

clock_t startTime;

double getCurrentTime() {
    return (double) (clock() - startTime) / CLOCKS_PER_SEC;
}

typedef long long ll;
mt19937 rng(chrono::high_resolution_clock::now().time_since_epoch().count());
const double eps = 0.00001;
const int MOD = 1e9 + 7;
const int INF = 1000000101;
const long long LLINF = 1223372000000000555;
const int N = 1e6 + 3e2;
const int M = 5555;

int p[N];
int a[N], b[N];
void solve(int TC) {
    int n;
    cin >> n;

    set<int> mn, mx;
    for (int i = 1; i <= n; i++){
        cin >> p[i];
        mn.insert(i);
        mx.insert(i);
    }

    for (int i = 1; i <= n; i++){
        if (p[i] != p[i - 1]){
            a[i] = b[i] = p[i];
            mn.erase(p[i]);
            mx.erase(p[i]);
        }
    }

    for (int i = 1; i <= n; i++){
        if (p[i] != p[i - 1]) continue;

        int x = *mn.begin(), y = *prev(mx.upper_bound(p[i]));

        a[i] = x;
        b[i] = y;

        mn.erase(x);
        mx.erase(y);
    }

    for (int i = 1; i <= n; i++) cout << a[i] << ' '; cout << endl;
    for (int i = 1; i <= n; i++) cout << b[i] << ' '; cout << endl;

}

int main() {
    startTime = clock();
    cin.tie(nullptr); cout.tie(nullptr);
    ios_base::sync_with_stdio(false);

    bool llololcal = false;
#ifdef dddxxz
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
    llololcal = true;
#endif

    int TC = 1;
    cin >> TC;

    for (int test = 1; test <= TC; test++) {
        debug(test);
        solve(test);
    }

    if (llololcal) cerr << endl << "Time: " << int(getCurrentTime() * 1000) << " ms" << endl;

    return 0;
}