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.util.*;
  import java.io.*;
  
  public class Main {
  	static class FastReader {
  		BufferedReader br;
  		StringTokenizer st;
  
  		public FastReader() {
  			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;
  		}
  	}
  
  	public static void sort(long[] a) {
  		ArrayList<Long> arr = new ArrayList<>();
  		for (int i = 0; i < a.length; i++)
  			arr.add(a[i]);
  		Collections.sort(arr);
  		for (int i = 0; i < arr.size(); i++)
  			a[i] = arr.get(i);
  	}
  
  	public static void revsort(long[] a) {
  		ArrayList<Long> arr = new ArrayList<>();
  		for (int i = 0; i < a.length; i++)
  			arr.add(a[i]);
  		Collections.sort(arr, Collections.reverseOrder());
  		for (int i = 0; i < arr.size(); i++)
  			a[i] = arr.get(i);
  	}
  
  	//Cover the small test cases like for n=1 .
  
  	public static void main(String[] args) {
  		FastReader obj = new FastReader();
  		PrintWriter out = new PrintWriter(System.out);
  		int len = obj.nextInt();
  		while (len-- != 0) {
  			int n = obj.nextInt();
  			int m=obj.nextInt();
  			int k=obj.nextInt();
  			int ceil=(int)Math.ceil((double)n/(double)m);
  			int floor=n/m;
              int L=n-m*floor;
              int R=m-L;
              int par=1;
              while(k--!=0)
              {
              	
              	for(int j=0;j<L;j++) {
              	 out.print(ceil+" ");
              	 for(int i=par;i<par+ceil;i++)
              	 {
              		int val=i%n;
              		if(val==0)val=n;
               		out.print(val+" ");
              	 }
                   par+=ceil;
              	 out.println();
              	}
              	int cat=par;
              	for(int j=0;j<R;j++)
              	{
              		out.print(floor+" ");
              		for(int i=cat;i<cat+floor;i++)
              		{
              			int val=i%n;
                  		if(val==0)val=n;
                   		out.print(val+" ");
              		}
              		cat+=floor;
                 	    out.println();
              	}
              }
              out.println();
  		}
  		out.flush();
  	}
  
  }