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:
	/*	/ ﾌﾌ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ム
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		(⠀ー '_人`ー ﾉ
		⠀|\ ￣ _人'彡ﾉ
		⠀ )\⠀⠀ ｡⠀⠀ /
		⠀⠀(\⠀ #⠀ /
		⠀/⠀⠀⠀/ὣ====================D-
		/⠀⠀⠀/⠀ \ \⠀⠀\
		( (⠀)⠀⠀⠀⠀ ) ).⠀)
		(⠀⠀)⠀⠀⠀⠀⠀( | /
		|⠀ /⠀⠀⠀⠀⠀⠀ | /
		[_] ⠀⠀⠀⠀⠀[___]                     */
	// Main Code at the Bottom
	import java.util.*;
	import java.lang.*;
	import java.io.*;
	import java.math.BigInteger; 
	public class Main {
		//Fast IO class
	    static class FastReader {
	        BufferedReader br; 
	        StringTokenizer st; 
	        public FastReader() {
	        	boolean env=System.getProperty("ONLINE_JUDGE") != null;
	        	if(!env) {
	        		try {
						br=new BufferedReader(new FileReader("src\\input.txt"));
					} catch (FileNotFoundException e) {
						e.printStackTrace();
					}
	        	}
	        	else br = new BufferedReader(new InputStreamReader(System.in)); 
	        } 
	        String next() {
	            while (st == null || !st.hasMoreElements()) {
	                try {
	                    st = new StringTokenizer(br.readLine()); 
	                } 
	                catch (IOException  e) {
	                    e.printStackTrace(); 
	                } 
	            } 
	            return st.nextToken(); 
	        } 
	        int nextInt() {
	            return Integer.parseInt(next()); 
	        } 
	        long nextLong() {
	            return Long.parseLong(next()); 
	        } 
	        double nextDouble() {
	            return Double.parseDouble(next()); 
	        } 
	        String nextLine() {
	            String str = ""; 
	            try {
	                str = br.readLine(); 
	            } 
	            catch (IOException e) {
	                e.printStackTrace(); 
	            } 
	            return str; 
	        } 
	    }     
	    static long MOD=1000000000+7;
	    //Euclidean Algorithm
	    static long gcd(long A,long B){
	    	return (B==0)?A:gcd(B,A%B);
	    }
	    //Modular Exponentiation
	    static long fastExpo(long x,long n){
	        if(n==0) return 1;
	        if((n&1)==0) return fastExpo((x*x)%MOD,n/2)%MOD;
	        return ((x%MOD)*fastExpo((x*x)%MOD,(n-1)/2))%MOD;
	    }
	    //Modular Inverse
	    static long inverse(long x) {
	    	return fastExpo(x,MOD-2);
	    }
	    //Prime Number Algorithm
	    static boolean isPrime(long n){
	        if(n<=1) return false;
	        if(n<=3) return true;
	        if(n%2==0 || n%3==0) return false;
	        for(int i=5;i*i<=n;i+=6) if(n%i==0 || n%(i+2)==0) return false;
	        return true;
	    }
	    //Reverse an array
	    static void reverse(char arr[],int l,int r){
	    	while(l<r) {
	    		char tmp=arr[l];
	    		arr[l++]=arr[r];
	    		arr[r--]=tmp;
	    	}
	    }
	    //Print array
	    static void print1d(int arr[]) {
	    	out.println(Arrays.toString(arr));
	    }
	    static void print2d(int arr[][]) {
	    	for(int a[]: arr) out.println(Arrays.toString(a));
	    }
	    // Pair
	    static class pair{
	    	int x,y;
	    	pair(int a,int b){
	    		this.x=a;
	    		this.y=b;
	    	}
	    	public boolean equals(Object obj) {
	    		if(obj == null || obj.getClass()!= this.getClass()) return false;
	            pair p = (pair) obj;
	            return (this.x==p.x && this.y==p.y);
	        }
	    	public int hashCode() {
	            return Objects.hash(x,y);
	        }
	    }
	    static FastReader sc=new FastReader();
	    static PrintWriter out=new PrintWriter(System.out);  
	    //Main function(The main code starts from here
	    public static void main (String[] args) throws java.lang.Exception {
	    	int test=1;
	    	test=sc.nextInt();
	    	while(test-->0){
	    		int n=sc.nextInt();
	    		TreeSet<Integer> a[]=new TreeSet[n-1];
	    		for(int i=0;i<n-1;i++) {
	    			int k=sc.nextInt();
	    			a[i]=new TreeSet<>();
	    			for(int j=0;j<k;j++) a[i].add(sc.nextInt());
	    		}
	    		ArrayList<Integer> ans=new ArrayList<>();
	    		for(int x=1;x<=n;x++) {
	    			int val=x,cnt=1;
	    			ArrayList<Integer> p=new ArrayList<>();
	    			p.add(val);
	    			TreeSet<Integer> tmp[]=new TreeSet[n-1];
	    			for(int i=0;i<n-1;i++) {
	    				tmp[i]=new TreeSet<>();
	    				for(Integer v:a[i]) tmp[i].add(v);
	    			}
	    			while(true) {
	    				int f=0, ne=val;
	    				for(int i=0;i<n-1;i++) {
	    					tmp[i].remove(val);
	    					if(tmp[i].size()==1) {
	    						f++;
	    						ne=tmp[i].first();
	    					}
	    				}
	    				if(f!=1) break;
	    				val=ne;
	    				p.add(val);
	    				cnt++;
	    				if(cnt==n) break;
	    			}
	    			if(cnt==n) {
	    				int flag=0;
	    				for(int i=0;i<n-1;i++) {
	    					int k=a[i].size(),fl=0;
	    					TreeSet<Integer> window=new TreeSet<>();
	    					for(int j=0;j<k;j++) window.add(p.get(j));
	    					if(window.equals(a[i])) fl=1;
	    					for(int j=k;j<n;j++) {
	    						window.remove(p.get(j-k));
	    						window.add(p.get(j));
	    						if(window.equals(a[i])) fl=1;
	    					}
	    					if(fl==0) {
	    						flag=1;
	    						break;
	    					}
	    				}
	    				if(flag==0) {
	    					ans=p;
	    					break;
	    				}
	    			}
	    		}
	    		for(Integer x: ans) out.print(x+" ");
	    		out.println();
	    	}
	        out.flush();
	        out.close();
	    }
	}