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;
#define ll long long
#define fast_io ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define endl "\n"
#define all(v) v.begin(),v.end()

const int mod = 1e9 + 7;
vector<int> getMin (int res[], set<int> s, int n) {
    vector<int> v(n);
    for (int i = 0; i < n; i++) {
        if (res[i]) {
            v[i] = res[i];
        }
        else {
            v[i] = *s.begin();
            s.erase(s.begin());
        }
    }
    return v;
}

vector<int> getMax (int res[], int a[], set<int> s, int n) {
    int sufMin[n];
    sufMin[n - 1] = n;
    for (int i = n - 2; i >= 0; i--) {
        if (res[i] != 0) {
            sufMin[i] = min(sufMin[i + 1], res[i]);
        }
        else sufMin[i] = sufMin[i + 1];
    }
    vector<int> v(n);
    for (int i = 0; i < n; i++) {
        if (res[i] == 0) {
            auto it = s.lower_bound(sufMin[i]);
            while (it == s.end() || (*it > sufMin[i] || *it > a[i])) {
                it--;
            }
            v[i] = *it;
            s.erase(v[i]);
        }
        else v[i] = res[i];
    }
    return v;
}
int main (){
    fast_io;
    int t;
    cin >> t;
    while (t--) {
        int n;
        cin >> n;
        int a[n];
        for (int i = 0; i < n; i++)
            cin >> a[i];
        int res[n];
        set<int> s;
        for (int i = 1; i <= n; i++)
            s.insert(i);
        memset(res, 0, sizeof(res));
        res[0] = a[0];
        s.erase(a[0]);
        for (int i = 1; i < n; i++) {
            if (a[i] != a[i - 1]) {
                res[i] = a[i];
                s.erase(a[i]);
            }
        }
        vector<int> mini = getMin(res, s, n);
        vector<int> maxi = getMax(res, a, s, n);
        for (auto it: mini)
            cout << it << ' ';
        cout << endl;
        for (auto it: maxi)
            cout << it << ' ';
        cout << endl;
    }
}