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.util.Map.Entry;
  
  import java.math.*;
  import java.io.*;
  
  public class Main {
  	public static void main(String[] args) throws FileNotFoundException {
  		InputReader in = new InputReader(System.in);
  		// Scanner in = new Scanner(System.in);
  		// Scanner in = new Scanner(new BufferedReader(new
  		// InputStreamReader(System.in)));
  		PrintWriter out = new PrintWriter(System.out);
  		// InputReader in = new InputReader(new
  		// File("ethan_traverses_a_tree.txt"));
  		// PrintWriter out = new PrintWriter(new
  		// File("ethan_traverses_a_tree-output.txt"));
  
  		int pi = in.nextInt();
  		for (int qi = 0; qi < pi; qi++) {
  			int n = in.nextInt();
  			int m = in.nextInt();
  			int k = in.nextInt();
  
  			int spe = n % m;
  			int tablenormalcount = n / m;
  			int tablespecount = tablenormalcount + 1;
  			int nowspeperson = 0;
  
  			for (int i = 0; i < k; i++) {
  				boolean[] used = new boolean[n];
  				List<List<Integer>> list = new ArrayList<List<Integer>>();
  				for (int j = 0; j < m; j++) {
  					list.add(new ArrayList<Integer>());
  				}
  
  				for (int j = m - spe; j < m; j++) {
  					for (int u = 0; u < tablespecount; u++) {
  						list.get(j).add(nowspeperson);
  						used[nowspeperson] = true;
  						nowspeperson = (nowspeperson + 1) % n;
  					}
  				}
  
  				int nowtable = 0;
  				for (int j = 0; j < n; j++) {
  					if (used[j] == false) {
  						list.get(nowtable).add(j);
  						nowtable = (nowtable + 1) % (m - spe);
  					}
  				}
  
  				for (int j = 0; j < m; j++) {
  					List<Integer> tablelist = list.get(j);
  					out.printf("%d", tablelist.size());
  					for (int u = 0; u < tablelist.size(); u++) {
  						out.printf(" %d", tablelist.get(u) + 1);
  					}
  					out.printf("\n");
  				}
  			}
  
  			if (qi != pi - 1) {
  				out.printf("\n");
  			}
  		}
  
  		out.close();
  	}
  
  	static class InputReader {
  		BufferedReader br;
  		StringTokenizer st;
  
  		public InputReader(File f) {
  			try {
  				br = new BufferedReader(new FileReader(f));
  			} catch (FileNotFoundException e) {
  				e.printStackTrace();
  			}
  		}
  
  		public InputReader(InputStream in) {
  			br = new BufferedReader(new InputStreamReader(in));
  		}
  
  		public String next() {
  			while (st == null || !st.hasMoreTokens()) {
  				try {
  					st = new StringTokenizer(br.readLine());
  				} catch (IOException e) {
  					e.printStackTrace();
  				}
  			}
  			return st.nextToken();
  		}
  
  		public boolean hasNext() {
  			while (st == null || !st.hasMoreTokens()) {
  				String s = null;
  				try {
  					s = br.readLine();
  				} catch (IOException e) {
  					e.printStackTrace();
  				}
  				if (s == null)
  					return false;
  				st = new StringTokenizer(s);
  			}
  			return true;
  		}
  
  		public int nextInt() {
  			return Integer.parseInt(next());
  		}
  
  		public long nextLong() {
  			return Long.parseLong(next());
  		}
  
  		public double nextDouble() {
  			return Double.parseDouble(next());
  		}
  	}
  }