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 FAST_IO  ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define show(x)  cerr << #x << " = " << x << '\n'
#define endl     '\n'
typedef long long ll;
typedef unsigned long long ull;
const double EPS = 1e-10;  // epsilon
const double PI = acos(-1);
const ll MOD = (ll)1e9 + 7;
const int N = 2e5 + 5;

int t = 1, n, a[N];
set<int> st;

vector<int> minimalPermutation();
vector<int> maximalPermutation();

int main()
{
    FAST_IO
    scanf("%d", &t);
    while (t--)
    {
        scanf("%d", &n);
        for (int i = 0; i < n; i++)
            scanf("%d", &a[i]);
        auto ans1 = minimalPermutation();
        auto ans2 = maximalPermutation();
        for(auto x : ans1)cout << x << ' '; cout << "\n";
        for(auto x : ans2)cout << x << ' '; cout << "\n";
    }
    return 0;
}

vector<int> minimalPermutation()
{
    st.clear();
    for (int i = 1; i <= n; i++)
        st.insert(i);
    vector<int> ans;
    ans.push_back(a[0]);
    st.erase(a[0]);
    for (int i = 1; i < n; i++)
    {
        if(a[i] != a[i - 1])
        {
            st.erase(a[i]);
            ans.push_back(a[i]);
        }
        else 
        {
            auto it = st.begin();
            ans.push_back(*it);
            st.erase(it);
        }
    }
    return ans;
}

vector<int> maximalPermutation()
{
    st.clear();
    for(int i = 1; i <= n; i++)
        st.insert(i);
    vector<int> ans;
    ans.push_back(a[0]);
    st.erase(a[0]);
    for (int i = 1; i < n; i++)
    {
        if (a[i] != a[i - 1]) 
        {
            st.erase(a[i]);
            ans.push_back(a[i]);
        }
        else 
        {
            auto it = --st.lower_bound(ans.back());
            ans.push_back(*it);
            st.erase(it);
        }
    }
    return ans;
}