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 nl '\n'
typedef long long ll;

void solve() {
    int n;
    cin >> n;
    int q[n], p_max[n], curr = -1, j = 0;
    queue<int> rev;
    set<int> permutation, rev_permutation;
    for (int i = 0; i < n; i++)
        permutation.insert(i+1), rev_permutation.insert(i+1);
    for (int i=0; i<n; i++) {
        cin >> q[i];
        permutation.erase(q[i]);
        rev_permutation.erase(q[i]);
    }
    for (int i=0; i<=n-1; i++) {
        if(q[i] != curr) {
            curr = q[i];
            cout << q[i] << " ";
            while(rev.size()) {
                p_max[j++] = rev.front();
                rev.pop();
            }
            p_max[j++] = q[i];
        } else {
            cout << *permutation.begin() << " ";
            auto it = rev_permutation.lower_bound(curr);
            it--;
            rev.push(*it);
            permutation.erase(permutation.begin());
            rev_permutation.erase(it);
        }
    }
    while(rev.size() && j<n) {
        p_max[j++] = rev.front();
        rev.pop();
    }
    p_max[j] = curr;
    cout << nl;
    for (int i=0; i<n; i++) cout << p_max[i] << " ";
    cout << nl;
}

int main() {
    ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
    int testCases;
    cin >> testCases;
    while(testCases--)
        solve();
    return 0;
}