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.BufferedReader;
  import java.io.IOException;
  import java.io.InputStreamReader;
  import java.io.PrintWriter;
  import java.util.StringTokenizer;
  
  public class CF_1619F {
  	public static void main(String[] args) throws NumberFormatException, IOException {
  		MyReader reader = new MyReader();
  		int t = reader.nextInt();
  		for(int testcase=0; testcase<t; testcase++) {
  			int n = reader.nextInt();
  			int m = reader.nextInt();
  			int k = reader.nextInt();
  			
  			int numBigTables = n - m*(n/m);
  			int smallTableSize = n/m;
  			int bigTableSize = smallTableSize + 1;
  			int totalPeopleAtBigTables = bigTableSize * numBigTables;
  			int sumbi = k * totalPeopleAtBigTables;
  			int smallerbiVal = sumbi/n;
  			int largerbiVal = smallerbiVal + 1;
  			int numLargerbi = sumbi - n*smallerbiVal;
  			
  			int[] arr = new int[n];
  			for(int i=0; i<numLargerbi; i++) {
  				arr[i] = largerbiVal;
  			}
  			for(int i=numLargerbi; i<n; i++) {
  				arr[i] = smallerbiVal;
  			}
  			
  			int[][] peopleAtBigTables = new int[k][totalPeopleAtBigTables];
  			for(int i=0; i<k; i++) {
  				int currSmaller = Integer.MAX_VALUE;
  				for(int ele : arr) {
  					currSmaller = Math.min(currSmaller, ele);
  				}
  				int currLarger = currSmaller+1;
  				int peopleChosenThisRound = 0;
  				for(int j=0; j<n; j++) {
  					if(arr[j] == currLarger) {
  						peopleAtBigTables[i][peopleChosenThisRound] = j;
  						peopleChosenThisRound++;
  						if(peopleChosenThisRound == totalPeopleAtBigTables) break;
  					}
  				}
  				if(peopleChosenThisRound != totalPeopleAtBigTables) {
  					for(int j=0; j<n; j++) {
  						if(arr[j] == currSmaller) {
  							peopleAtBigTables[i][peopleChosenThisRound] = j;
  							peopleChosenThisRound++;
  							if(peopleChosenThisRound == totalPeopleAtBigTables) break;
  						}
  					}
  				}
  				
  				for(int personChosen : peopleAtBigTables[i]) {
  					arr[personChosen]--;
  				}
  			}
  			
  			//output
  			PrintWriter pw = new PrintWriter(System.out);
  			for(int i=0; i<k; i++) {
  				boolean[] used = new boolean[n];
  				for(int j=0; j<peopleAtBigTables[i].length; j++) {
  					//every j: new line, so print # elements on this line
  					if(j % bigTableSize == 0) {
  						pw.println();
  						pw.print(bigTableSize + " ");
  					}
  					
  					pw.print((peopleAtBigTables[i][j]+1) + " ");
  					used[peopleAtBigTables[i][j]] = true;
  				}
  				
  				int lineCount = 0;
  				for(int person=0; person<n; person++) {
  					if(!used[person]) {
  						if(lineCount == 0) {
  							pw.println();
  							pw.print((bigTableSize-1) + " ");
  						}
  						pw.print((person+1) + " ");
  						lineCount++;
  						if(lineCount == bigTableSize-1) {
  							lineCount = 0;
  						}
  					}
  				}
  			}
  			pw.flush();
  			
  		}
  	}
  	
  	static class MyReader {
          BufferedReader br;
          StringTokenizer st;
          public MyReader() {
              br = new BufferedReader(new InputStreamReader(System.in));
          }
          String next() throws IOException  { 
          	while (st == null || !st.hasMoreElements()) { 
              	st = new StringTokenizer(br.readLine());
          	} 
          	return st.nextToken(); 
          } 
          int nextInt() throws NumberFormatException, IOException { 
          	return Integer.parseInt(next()); 
          }
          long nextLong() throws NumberFormatException, IOException { 
          	return Long.parseLong(next()); 
          }
          double nextDouble() throws NumberFormatException, IOException { 
              return Double.parseDouble(next()); 
          }
          String nextLine() throws IOException {
              return br.readLine(); 
          } 
      }
  }
  