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")
#pragma GCC optimization ("unroll-loops")
 
#include <bits/stdc++.h>
 
#define IOS ios_base::sync_with_stdio(0); cin.tie(0);
#define pb push_back
#define pf push_front
#define fi first
#define se second
#define ins insert
#define sz(x) (int)x.size()
#define dbg(x) cout << x << "\n";
 
const int N = 2e5 + 5;
const long long int mod = 1e9 + 7;
const long long int Mod = 998244353;
const long double Pi = acos(-1);
const long long int Inf = 4e18;
 
using namespace std;
 
void TestCase (){
        int n;
        cin >> n;
        vector <int> q(n), mi(n), ma(n);
        for(int i = 0; i<n; i++) cin >> q[i];
        mi[0] = ma[0] = q[0];
        // minimum
        set <int> st;
        for(int i = 1; i<=n; i++) st.ins(i);
        st.erase(q[0]);
        for(int i = 1; i<n; i++){
                if(q[i] == q[i - 1]){
                        mi[i] = *st.begin();
                        st.erase(st.begin());
                }
                else{
                        mi[i] = q[i];
                        st.erase(q[i]);
                }
        }
        // maximum
        for(int i = 1; i<=n; i++) st.ins(i);
        st.erase(q[0]);
        int last = q[0];
        for(int i = 1; i<n; i++){
                if(q[i] == q[i - 1]){
                        auto low = st.lower_bound(last);
                        low--;
                        ma[i] = *low;
                        last = ma[i];
                        st.erase(low);
                }
                else{
                        ma[i] = q[i];
                        last = ma[i];
                        st.erase(ma[i]);
                }
        }
        for(int i = 0; i<n; i++) cout << mi[i] << " ";
        cout << "\n";
        for(int i = 0; i<n; i++) cout << ma[i] << " ";
}
 
int main(){
        IOS;
        int T = 1;
        cin >> T;
        while(T--){
                TestCase();
                cout << "\n";
        }
        return 0;
}