description: The Hat is a game of speedy explanation/guessing words (similar to Alias). It's fun. Try it! In this problem, we are talking about a variant of the game when the players are sitting at the table and everyone plays individually (i.e. not teams, but individual gamers play).
  
  n people gathered in a room with m tables (n ≥ 2m). They want to play the Hat k times. Thus, k games will be played at each table. Each player will play in k games.
  
  To do this, they are distributed among the tables for each game. During each game, one player plays at exactly one table. A player can play at different tables.
  
  Players want to have the most "fair" schedule of games. For this reason, they are looking for a schedule (table distribution for each game) such that:
  
    * At any table in each game there are either ⌊n/m⌋ people or ⌈n/m⌉ people (that is, either n/m rounded down, or n/m rounded up). Different numbers of people can play different games at the same table.
    * Let's calculate for each player the value b_i — the number of times the i-th player played at a table with ⌈n/m⌉ persons (n/m rounded up). Any two values of b_imust differ by no more than 1. In other words, for any two players i and j, it must be true |b_i - b_j| ≤ 1. 
  
  
  
  For example, if n=5, m=2 and k=2, then at the request of the first item either two players or three players should play at each table. Consider the following schedules:
  
    * First game: 1, 2, 3 are played at the first table, and 4, 5 at the second one. The second game: at the first table they play 5, 1, and at the second — 2, 3, 4. This schedule is not "fair" since b_2=2 (the second player played twice at a big table) and b_5=0 (the fifth player did not play at a big table).
    * First game: 1, 2, 3 are played at the first table, and 4, 5 at the second one. The second game: at the first table they play 4, 5, 2, and at the second one — 1, 3. This schedule is "fair": b=[1,2,1,1,1] (any two values of b_i differ by no more than 1). 
  
  
  
  Find any "fair" game schedule for n people if they play on the m tables of k games.
  
  Input
  
  The first line of the input contains an integer t (1 ≤ t ≤ 10^4) — the number of test cases in the test.
  
  Each test case consists of one line that contains three integers n, m and k (2 ≤ n ≤ 2⋅10^5, 1 ≤ m ≤ ⌊n/2⌋, 1 ≤ k ≤ 10^5) — the number of people, tables and games, respectively.
  
  It is guaranteed that the sum of nk (n multiplied by k) over all test cases does not exceed 2⋅10^5.
  
  Output
  
  For each test case print a required schedule — a sequence of k blocks of m lines. Each block corresponds to one game, a line in a block corresponds to one table. In each line print the number of players at the table and the indices of the players (numbers from 1 to n) who should play at this table.
  
  If there are several required schedules, then output any of them. We can show that a valid solution always exists.
  
  You can output additional blank lines to separate responses to different sets of inputs.
  
  Example
  
  Input
  
  
  3
  5 2 2
  8 3 1
  2 1 3
  
  
  Output
  
  
  3 1 2 3
  2 4 5
  3 4 5 2
  2 1 3
  
  2 6 2
  3 3 5 1
  3 4 7 8
  
  2 2 1
  2 2 1
  2 2 1
solution: import java.io.*;
  
  import java.util.*;
  /*
  
  
  
  1
  5 2 2
  
  */
  
   
   public class F{
  	static FastReader sc=null;
  	
  	public static void main(String[] args) {
  		sc=new FastReader();
  		int t=sc.nextInt();
  		PrintWriter out=new PrintWriter(System.out);
  		
  		for(int tt=0;tt<t;tt++) {
  			int n=sc.nextInt(),m=sc.nextInt(),k=sc.nextInt();
  			//r = no of extras each game = n%m
  			//(k*r)/n = for each of them no of times it is extra
  			
  			int extra_tables=n%m,each=(n+m-1)/m,tot=each*extra_tables;
  			
  			//System.out.println(tot+" "+extra_tables+" "+each);
  			ArrayList<Integer> extras[]=new ArrayList[k];
  			for(int i=0;i<k;i++) extras[i]=new ArrayList<>();
  			
  			for(int i=0;i<k;i++) {
  				int cur=(i*tot)%n;
  				for(int j=cur;j<cur+tot;j++)extras[i].add((j%n));
  			}
  			
  			//for(int g=0;g<k;g++)System.out.println(extras[g].toString());
  			
  			
  			for(int g=0;g<k;g++) {
  				//split extras into n/m each
  				boolean taken[]=new boolean[n];
  				int cur=0;
  				for(int i=0;i<extra_tables;i++) {
  					out.print(each+" ");
  					for(int j=0;j<each;j++) {
  						int val=extras[g].get(cur++);
  						taken[val]=true;
  						out.print((val+1)+" ");
  					}
  					out.println();
  				}
  				cur=0;
  				int pop=each;
  				if(extra_tables>0)pop=each-1;
  				
  				for(int i=extra_tables;i<m;i++) {
  					out.print(pop+" ");
  					for(int j=0;j<pop;j++) {
  						while(taken[cur]) {
  							cur++;
  							cur%=n;
  						}
  						out.print((cur+1)+" ");
  						cur++;
  					}
  					out.println();
  				}
  			}
  			out.println();
  		}
  		
  		out.close();
  	}
  	
  	static int[] ruffleSort(int a[]) {
  		ArrayList<Integer> al=new ArrayList<>();
  		for(int i:a)al.add(i);
  		Collections.sort(al);
  		for(int i=0;i<a.length;i++)a[i]=al.get(i);
  		return a;
  	}
  	
  	static void print(int a[]) {
  		for(int e:a) {
  			System.out.print(e+" ");
  		}
  		System.out.println();
  	}
  	
  	static class FastReader{
  		
  		StringTokenizer st=new StringTokenizer("");
  		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
  		
  		String next() {
  			while(!st.hasMoreTokens()) 
  				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());
  		}
  		
  		int[] readArray(int n) {
  			int a[]=new int[n];
  			for(int i=0;i<n;i++)a[i]=sc.nextInt();
  			return a;
  		}
  	}
  	
  	
  }
  