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.*;
import java.util.*;

public class Main{
	static class segment implements Comparable<segment>{
		Integer[]nums;
		segment(Integer[]in){
			nums=in.clone();
		}
		@Override
		public int compareTo(segment o) {
			if(nums.length!=o.nums.length)return nums.length-o.nums.length;
			
			for(int i=0;i<nums.length;i++) {
				if(nums[i]<o.nums[i])return -1;
				if(nums[i]>o.nums[i])return 1;
			}
			return 0;
		}
		public String toString() {
			return Arrays.toString(nums);
		}
	}
	public static void main(String[] args) throws Exception{
		PrintWriter pw=new PrintWriter(System.out);
		MScanner sc = new MScanner(System.in);
		int tc=sc.nextInt();
		while(tc-->0) {
			int n=sc.nextInt();
			int[][]in=new int[n-1][];
			for(int i=0;i<n-1;i++) {
				in[i]=sc.intArr(sc.nextInt());
			}
			for(int start=1;start<=n;start++) {
				HashSet<Integer>[]segments=new HashSet[n-1];
				for(int i=0;i<n-1;i++) {
					segments[i]=new HashSet<>();
					for(int j:in[i])segments[i].add(j);
				}
				int cur=start;
				LinkedList<Integer>ans=new LinkedList<>();
				boolean ok=true;
				zz:while(true) {
					ans.add(cur);
					for(int i=0;i<n-1;i++) {
						if(segments[i].contains(cur)) {
							segments[i].remove(cur);
						}
					}
					boolean empty=true;
					for(int i=0;i<n-1;i++) {
						if(segments[i].size()>0) {
							empty=false;
						}
					}
					if(empty) {
						break;
					}
					int idx=-1;
					for(int i=0;i<n-1;i++) {
						if(segments[i].size()==1) {
							if(idx!=-1) {
								ok=false;
								break zz;
							}
							idx=i;
						}
					}
					if(idx==-1) {
						ok=false;
						break;
					}
					for(int i:segments[idx]) {
						cur=i;
					}
					segments[idx].remove(cur);
				}
				if(!ok || ans.size()!=n)continue;
				int[]finalAns=new int[n];int c=0;
				while(!ans.isEmpty()) {
					finalAns[c++]=ans.pollFirst();
				}
				TreeSet<segment>ts=new TreeSet<segment>();
				for(int i=0;i<n-1;i++) {
					Integer[]x=new Integer[in[i].length];
					for(int j=0;j<x.length;j++) {
						x[j]=in[i][j];
					}
					ts.add(new segment(x));
				}
				for(int i=1;i<n;i++) {
					boolean found=false;
					for(int j=i-1;j>=0;j--) {
						Integer[]curSegment=new Integer[i-j+1];int cc=0;
						for(int k=j;k<=i;k++) {
							curSegment[cc++]=finalAns[k];							
						}
						Arrays.sort(curSegment);
						segment s=new segment(curSegment);
						if(ts.contains(s)) {
							ts.remove(s);
							found=true;
							break;
						}
					}
					if(!found) {
						ok=false;break;
					}
				}
				if(!ok)continue;
				for(int i:finalAns)pw.print(i+" ");
				pw.println();
				break;
			}
		}
		
		pw.flush();
	}
	static class MScanner {
		StringTokenizer st;
		BufferedReader br;
		public MScanner(InputStream system) {
			br = new BufferedReader(new InputStreamReader(system));
		}
 
		public MScanner(String file) throws Exception {
			br = new BufferedReader(new FileReader(file));
		}
 
		public String next() throws IOException {
			while (st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}
		public int[] intArr(int n) throws IOException {
	        int[]in=new int[n];for(int i=0;i<n;i++)in[i]=nextInt();
	        return in;
		}
		public long[] longArr(int n) throws IOException {
	        long[]in=new long[n];for(int i=0;i<n;i++)in[i]=nextLong();
	        return in;
		}
		public int[] intSortedArr(int n) throws IOException {
	        int[]in=new int[n];for(int i=0;i<n;i++)in[i]=nextInt();
	        shuffle(in);
	        Arrays.sort(in);
	        return in;
		}
		public long[] longSortedArr(int n) throws IOException {
	        long[]in=new long[n];for(int i=0;i<n;i++)in[i]=nextLong();
	        shuffle(in);
	        Arrays.sort(in);
	        return in;
		}
		public Integer[] IntegerArr(int n) throws IOException {
	        Integer[]in=new Integer[n];for(int i=0;i<n;i++)in[i]=nextInt();
	        return in;
		}
		public Long[] LongArr(int n) throws IOException {
	        Long[]in=new Long[n];for(int i=0;i<n;i++)in[i]=nextLong();
	        return in;
		}
		public String nextLine() throws IOException {
			return br.readLine();
		}
 
		public int nextInt() throws IOException {
			return Integer.parseInt(next());
		}
 
		public double nextDouble() throws IOException {
			return Double.parseDouble(next());
		}
 
		public char nextChar() throws IOException {
			return next().charAt(0);
		}
 
		public long nextLong() throws IOException {
			return Long.parseLong(next());
		}
 
		public boolean ready() throws IOException {
			return br.ready();
		}
 
		public void waitForInput() throws InterruptedException {
			Thread.sleep(3000);
		}
	}
	static void shuffle(int[]in) {
		for(int i=0;i<in.length;i++) {
			int idx=(int)(Math.random()*in.length);
			int tmp=in[i];
			in[i]=in[idx];
			in[idx]=tmp;
		}
	}
	static void shuffle(long[]in) {
		for(int i=0;i<in.length;i++) {
			int idx=(int)(Math.random()*in.length);
			long tmp=in[i];
			in[i]=in[idx];
			in[idx]=tmp;
		}
	}
}