Problem: We guessed a permutation p consisting of n integers. The permutation of length n is the array of length n where each element from 1 to n appears exactly once. This permutation is a secret for you.

For each position r from 2 to n we chose some other index l (l < r) and gave you the segment p_l, p_{l + 1}, ..., p_r in sorted order (i.e. we rearranged the elements of this segment in a way that the elements of this segment are sorted). Thus, you are given exactly n-1 segments of the initial permutation but elements inside each segment are sorted. The segments are given to you in random order.

For example, if the secret permutation is p=[3, 1, 4, 6, 2, 5] then the possible given set of segments can be:

  * [2, 5, 6] 
  * [4, 6] 
  * [1, 3, 4] 
  * [1, 3] 
  * [1, 2, 4, 6] 



Your task is to find any suitable permutation (i.e. any permutation corresponding to the given input data). It is guaranteed that the input data corresponds to some permutation (i.e. such permutation exists).

You have to answer t independent test cases.

Input

The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases. Then t test cases follow.

The first line of the test case contains one integer n (2 ≤ n ≤ 200) — the length of the permutation.

The next n-1 lines describe given segments.

The i-th line contains the description of the i-th segment. The line starts with the integer k_i (2 ≤ k_i ≤ n) — the length of the i-th segment. Then k_i integers follow. All integers in a line are distinct, sorted in ascending order, between 1 and n, inclusive.

It is guaranteed that the required p exists for each test case.

It is also guaranteed that the sum of n over all test cases does not exceed 200 (∑ n ≤ 200).

Output

For each test case, print the answer: n integers p_1, p_2, ..., p_n (1 ≤ p_i ≤ n, all p_i should be distinct) — any suitable permutation (i.e. any permutation corresponding to the test case input).

Example

Input


5
6
3 2 5 6
2 4 6
3 1 3 4
2 1 3
4 1 2 4 6
5
2 2 3
2 1 2
2 1 4
2 4 5
7
3 1 2 6
4 1 3 5 6
2 1 2
3 4 5 7
6 1 2 3 4 5 6
3 1 3 6
2
2 1 2
5
2 2 5
3 2 3 5
4 2 3 4 5
5 1 2 3 4 5


Output


3 1 4 6 2 5 
3 2 1 4 5 
2 1 6 3 5 4 7 
1 2 
2 5 3 4 1 

Solution (in JAVA):

Solution:
/*
If you want to aim high, aim high
Don't let that studying and grades consume you
Just live life young
******************************
If I'm the sun, you're the moon
Because when I go up, you go down
*******************************
I'm working for the day I will surpass you
https://www.a2oj.com/Ladder16.html
*/
import java.util.*;
import java.io.*;
import java.math.*;

   public class x1343F
   {
      public static void main(String omkar[]) throws Exception
      {
         BufferedReader infile = new BufferedReader(new InputStreamReader(System.in));  
         StringTokenizer st = new StringTokenizer(infile.readLine());
         int T = Integer.parseInt(st.nextToken());
         StringBuilder sb = new StringBuilder();
         outer:while(T-->0)
         {
            int N = Integer.parseInt(infile.readLine());
            ranges = new ArrayList[N-1];
            for(int i=0; i < N-1; i++)
            {
               st = new StringTokenizer(infile.readLine());
               int x = Integer.parseInt(st.nextToken());
               ranges[i] = new ArrayList<Integer>();
               for(int a=0; a < x; a++)
                  ranges[i].add(Integer.parseInt(st.nextToken()));
            }
            //1st one must be length 2
            boolean[] used = new boolean[N];
            int[] res = new int[N];
            for(int a=0; a < N-1; a++)
               if(ranges[a].size() == 2)
               {
                  res[0] = ranges[a].get(0);
                  res[1] = ranges[a].get(1);
                  used[a] = true;
                  if(dfs(res, used))
                  {
                     for(int x: res)
                        sb.append(x+" ");
                     sb.append("\n");
                     continue outer;
                  }
                  res[1] = ranges[a].get(0);
                  res[0] = ranges[a].get(1);
                  if(dfs(res, used))
                  {
                     for(int x: res)
                        sb.append(x+" ");
                     sb.append("\n");
                     continue outer;
                  }
                  used[a] = false;
               }
         }
         System.out.print(sb);
      }
      static ArrayList<Integer>[] ranges;
      public static boolean dfs(int[] res, boolean[] used)
      {
         int curr;
         for(curr=0; curr < res.length; curr++)
            if(res[curr] == 0)
               break;
         if(curr == res.length)
            return true;
         int N = res.length;
         for(int dex=0; dex < N-1; dex++)
         {
            if(used[dex])
               continue;
            ArrayList<Integer> ls = ranges[dex];
            if(ls.size() <= curr+1)
            {
               HashSet<Integer> temp = new HashSet<Integer>();
               for(int x: ls)
                  temp.add(x);
               for(int i=curr-1; i > curr-ls.size(); i--)
                  if(temp.contains(res[i]))
                     temp.remove(res[i]);
               if(temp.size() == 1)
               {
                  int added = 0;
                  for(int x: temp)
                     added = x;
                  if(!contained(res, added))
                  {
                     used[dex] = true;
                     res[curr] = added;
                     if(dfs(res, used))
                        return true;
                     res[curr] = 0;
                     used[dex] = false;
                  }
               }
            }
         }
         return false;
      }
      public static boolean contained(int[] res, int x)
      {
         for(int v: res)
            if(v == x)
               return true;
         return false;
      }
   }