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:
//package round636;
import java.io.ByteArrayInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.InputMismatchException;
import java.util.TreeSet;

public class F3 {
	InputStream is;
	PrintWriter out;
	String INPUT = "";
	
	void solve()
	{
		for(int T = ni();T > 0;T--)go();
	}
	
	void go()
	{
		int n = ni();
		int[][] a = new int[n-1][];
		for(int i = 0;i < n-1;i++){
			a[i] = na(ni());
		}
		
		int[] route;
		out:
		{
			int[] f = new int[n+1];
			int[] last = new int[n+1];
			for(int i = 0;i < n-1;i++){
				for(int v : a[i]){
					f[v]++;
					last[v] = i;
				}
			}
			int[] of = Arrays.copyOf(f, n+1);
			
			route = new int[n];
			outer:
			for(int i = 1;i <= n;i++){
				if(f[i] == 1){
					boolean[] done = new boolean[n];
					
					int cur = i;
					int s = n*(n+1)/2;
					for(int q = 0;q < n-2;q++){
						route[n-1-q] = cur;
						s -= cur;
						int can = -1;
//						tr(f, cur, last[cur], a[last[cur]]);
						for(int v : a[last[cur]]){
							if(f[v] == 2){
								if(can == -1 || of[v] > of[can]){
									can = v;
								}
							}
						}
						if(can == -1)continue outer;
						done[last[cur]] = true;
						cur = can;
						
						Arrays.fill(f, 0);
						for(int j = 0;j < n-1;j++){
							if(!done[j]){
								for(int v : a[j]){
									f[v]++;
									last[v] = j;
								}
							}
						}
					}
					route[n-1-(n-2)] = cur;
					s -= cur;
					route[0] = s;
					break out;
				}
			}
		}
//		check(route, a);
		
		for(int v : route){
			out.print(v + " ");
		}
		out.println();
	}
	
	void check(int[] a, int[][] qs)
	{
		int n = a.length;
		int ptn = 0;
		for(int v : a)ptn ^= 1<<v;
		assert ptn+2 == (1<<n+1);
		for(int[] q : qs)Arrays.sort(q);
		
		boolean[] done = new boolean[n-1];
		outer:
		for(int i = 0;i < n-1;i++){
			TreeSet<Integer> h = new TreeSet<>();
			for(int j = i+1;j >= 0;j--){
				h.add(a[j]);
				if(h.size() >= 2){
					inner:
					for(int k = 0;k < n-1;k++){
						if(done[k])continue;
						if(h.size() != qs[k].length)continue;
						int p = 0;
						for(int t : h){
							if(t != qs[k][p++]){
								continue inner;
							}
						}
						done[k] = true;
						continue outer;
					}
				}
			}
			tr(i, a, h);
			assert false;
		}
	}
	
	void run() throws Exception
	{
		is = oj ? System.in : new ByteArrayInputStream(INPUT.getBytes());
		out = new PrintWriter(System.out);
		
		long s = System.currentTimeMillis();
		solve();
		out.flush();
		tr(System.currentTimeMillis()-s+"ms");
	}
	
	public static void main(String[] args) throws Exception { new F3().run(); }
	
	private byte[] inbuf = new byte[1024];
	public int lenbuf = 0, ptrbuf = 0;
	
	private int readByte()
	{
		if(lenbuf == -1)throw new InputMismatchException();
		if(ptrbuf >= lenbuf){
			ptrbuf = 0;
			try { lenbuf = is.read(inbuf); } catch (IOException e) { throw new InputMismatchException(); }
			if(lenbuf <= 0)return -1;
		}
		return inbuf[ptrbuf++];
	}
	
	private boolean isSpaceChar(int c) { return !(c >= 33 && c <= 126); }
	private int skip() { int b; while((b = readByte()) != -1 && isSpaceChar(b)); return b; }
	
	private double nd() { return Double.parseDouble(ns()); }
	private char nc() { return (char)skip(); }
	
	private String ns()
	{
		int b = skip();
		StringBuilder sb = new StringBuilder();
		while(!(isSpaceChar(b))){ // when nextLine, (isSpaceChar(b) && b != ' ')
			sb.appendCodePoint(b);
			b = readByte();
		}
		return sb.toString();
	}
	
	private char[] ns(int n)
	{
		char[] buf = new char[n];
		int b = skip(), p = 0;
		while(p < n && !(isSpaceChar(b))){
			buf[p++] = (char)b;
			b = readByte();
		}
		return n == p ? buf : Arrays.copyOf(buf, p);
	}
	
	private char[][] nm(int n, int m)
	{
		char[][] map = new char[n][];
		for(int i = 0;i < n;i++)map[i] = ns(m);
		return map;
	}
	
	private int[] na(int n)
	{
		int[] a = new int[n];
		for(int i = 0;i < n;i++)a[i] = ni();
		return a;
	}
	
	private int ni()
	{
		int num = 0, b;
		boolean minus = false;
		while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'));
		if(b == '-'){
			minus = true;
			b = readByte();
		}
		
		while(true){
			if(b >= '0' && b <= '9'){
				num = num * 10 + (b - '0');
			}else{
				return minus ? -num : num;
			}
			b = readByte();
		}
	}
	
	private long nl()
	{
		long num = 0;
		int b;
		boolean minus = false;
		while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'));
		if(b == '-'){
			minus = true;
			b = readByte();
		}
		
		while(true){
			if(b >= '0' && b <= '9'){
				num = num * 10 + (b - '0');
			}else{
				return minus ? -num : num;
			}
			b = readByte();
		}
	}
	
	private boolean oj = System.getProperty("ONLINE_JUDGE") != null;
	private void tr(Object... o) { if(!oj)System.out.println(Arrays.deepToString(o)); }
}