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,unroll-loops")
//#pragma GCC target("avx,avx2,sse,sse2")
#include <bits/stdc++.h>
using namespace std;
#define rep(i, j, n) for (ll i = j; i <= n; i++)
#define per(i, j, n) for (ll i = j; i >= n; i--)
#define ll long long
#define ld long double
#define vi vector<ll>
constexpr ll mod = 1e9 + 7;

ll t = 1;
ll n;

void solve() {
  // Code goes here
  cin >> n;
  vi q(n);
  rep(i, 0, n - 1) cin >> q[i];
  ll minm = 1;
  vi lmin, lmax;
  set<ll> done;
  done.insert(q[0]);
  lmin.push_back(q[0]);
  rep(i, 1, n - 1) {
    if (q[i] == q[i - 1]) {
      while (done.count(minm))
        ++minm;
      lmin.push_back(minm);
      done.insert(minm);
    } else {
      lmin.push_back(q[i]);
      done.insert(q[i]);
    }
  }
  done.clear();
  rep(i, 0, n - 1) done.insert(i + 1);
  done.erase(q[0]);
  lmax.push_back(q[0]);
  rep(i, 1, n - 1) {
    if (q[i] == q[i - 1]) {
      ll cur = q[i] - 1;
      auto it = done.upper_bound(q[i]);
      --it;
      lmax.push_back(*it);
      done.erase(it);
    } else {
      lmax.push_back(q[i]);
      done.erase(q[i]);
    }
  }
  rep(i, 0, n - 1) cout << lmin[i] << ' ';
  cout << '\n';
  rep(i, 0, n - 1) cout << lmax[i] << ' ';
  cout << '\n';
}

int main() {
  ios::sync_with_stdio(false);
  cin >> t;
  rep(i, 1, t) {
    // cout<<"Case #" << i << ": ";
    solve();
  }
}