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:
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Iterator;
import java.util.InputMismatchException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.HashSet;
import java.util.StringTokenizer;
import java.io.BufferedReader;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
	static class TaskAdapter implements Runnable {
		@Override
		public void run() {
			InputStream inputStream = System.in;
			OutputStream outputStream = System.out;
			Input in = new Input(inputStream);
			PrintWriter out = new PrintWriter(outputStream);
			FRestoreThePermutationBySortedSegments solver = new FRestoreThePermutationBySortedSegments();
			int testCount = Integer.parseInt(in.next());
			for(int i = 1; i<=testCount; i++)
				solver.solve(i, in, out);
			out.close();
		}
	}

	public static void main(String[] args) throws Exception {
		Thread thread = new Thread(null, new TaskAdapter(), "", 1<<29);
		thread.start();
		thread.join();
	}
	static class FRestoreThePermutationBySortedSegments {
		public FRestoreThePermutationBySortedSegments() {
		}

		public void solve(int testNumber, Input in, PrintWriter pw) {
			int n = in.nextInt();
			HashSet<Integer>[] segs = new HashSet[n-1];
			for(int i = 0; i<n-1; i++) {
				segs[i] = new HashSet<>();
				int x = in.nextInt();
				for(int j = 0; j<x; j++) {
					segs[i].add(in.nextInt());
				}
			}
			loop:
			for(int first = 1; first<=n; first++) {
				HashSet<Integer>[] tmp = new HashSet[n-1];
				for(int i = 0; i<tmp.length; i++) {
					tmp[i] = new HashSet<>(segs[i]);
				}
				ArrayList<Integer> cur = new ArrayList<>();
				cur.add(first);
				for(int i = 0; i<n-1; i++) {
					int x = cur.get(i);
					ArrayList<HashSet<Integer>> size1 = new ArrayList<>();
					for(int j = 0; j<tmp.length; j++) {
						if(tmp[j].remove(x)&&tmp[j].size()==1) {
							size1.add(tmp[j]);
						}
					}
					if(size1.size()!=1) {
						continue loop;
					}
					cur.add(size1.get(0).iterator().next());
				}
				boolean[] marked = new boolean[n];
				inner:
				for(HashSet<Integer> hs: segs) {
					HashSet<Integer> curhs = new HashSet<>(hs);
					iner:
					for(int i = 0; i<=n-curhs.size(); i++) {
						if(marked[i+curhs.size()-1]) {
							continue;
						}
						for(int j = i; j<i+curhs.size(); j++) {
							if(!curhs.contains(cur.get(j))) {
								continue iner;
							}
						}
						marked[i+curhs.size()-1] = true;
						continue inner;
					}
					continue loop;
				}
				int[] ans = new int[n];
				for(int i = 0; i<n; i++) {
					ans[i] = cur.get(i);
				}
				pw.println(Utilities.getString(ans));
				return;
			}
		}

	}

	static class Input {
		BufferedReader br;
		StringTokenizer st;

		public Input(InputStream is) {
			br = new BufferedReader(new InputStreamReader(is), 1<<16);
			st = null;
		}

		public boolean hasNext() {
			try {
				while(st==null||!st.hasMoreTokens()) {
					String s = br.readLine();
					if(s==null) {
						return false;
					}
					st = new StringTokenizer(s);
				}
				return true;
			}catch(Exception e) {
				return false;
			}
		}

		public String next() {
			if(!hasNext()) {
				throw new InputMismatchException();
			}
			return st.nextToken();
		}

		public int nextInt() {
			return Integer.parseInt(next());
		}

	}

	static class Utilities {
		public static String getString(int[] arr) {
			return getString(arr, "", " ", "");
		}

		public static String getString(int[] arr, String start, String middle, String end) {
			StringBuilder builder = new StringBuilder();
			builder.append(start);
			for(int i = 0; i<arr.length; i++) {
				if(i!=0) {
					builder.append(middle);
				}
				builder.append(arr[i]);
			}
			builder.append(end);
			return builder.toString();
		}

	}
}