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.lang.*;
  import java.io.*;
  public class Main
  {
      static void sort(int a[]){
          ArrayList<Integer> arr=new ArrayList<>();
          for(int i=0;i<a.length;i++)arr.add(a[i]);
          Collections.sort(arr);
          for(int i=0;i<a.length;i++)a[i]=arr.get(i);
      }
  	public static void main (String[] args) throws java.lang.Exception
  	{
  	    Scanner sc=new Scanner(System.in);
  	   // BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
  	    int t=1;
  	    t=sc.nextInt();
  	  //  int t=Integer.parseInt(br.readLine());
  	  PrintWriter out=new PrintWriter(System.out);
  	    while(--t>=0){
  	        int n=sc.nextInt();
  	        int m=sc.nextInt();
  	        int k=sc.nextInt();
  	        int z=(n+m-1)/m;
  	        int d=(n%m)*z;
  	        if(d==0)d=n;
  	        
  	        int dp[][]=new int[k+5][n+5];
  	        int ans[][]=new int[k+5][n+5];
  	        int max=1;
  	        for(int i=1;i<=k;i++){
  	            int cnt=0;
  	            boolean v[]=new boolean[n+5];
  	            for(int j=1;j<=n;j++){
  	                if((dp[i-1][j]+1)==max&&cnt<d){
  	                    cnt++;
  	                    ans[i][cnt]=j;
  	                    dp[i][j]=dp[i-1][j]+1;
  	                    //cnt++;
  	                    v[j]=true;
  	                }
  	            }
  	            for(int j=1;j<=n;j++){
  	                if(dp[i-1][j]==max&&cnt<d&&!v[j]){
  	                    cnt++;
  	                    ans[i][cnt]=j;
  	                    dp[i][j]=dp[i-1][j]+1;
  	                    //cnt++;
  	                    v[j]=true;
  	                }
  	            }
  	            for(int j=1;j<=n;j++){
  	                if(!v[j]){
  	                    cnt++;
  	                    ans[i][cnt]=j;
  	                 dp[i][j]=dp[i-1][j];
  	                }
  	            }
  	            for(int j=1;j<=n;j++)max=Math.max(max,dp[i][j]);
  	        }
  	        
  	        for(int i=1;i<=k;i++){
  	            for(int j=1;j<=d;j=j+z){
  	                out.print(z+" ");
  	                for(int l=j;l<j+z;l++){
  	                    out.print(ans[i][l]+" ");
  	                }
  	                out.println();
  	            }
  	            for(int j=d+1;j<=n;j=j+z-1){
  	                out.print((z-1)+" ");
  	                for(int l=j;l<j+z-1;l++){
  	                    out.print(ans[i][l]+" ");
  	                }
  	                out.println();
  	            }
  	            
  	        }
  	        
  	       
  	     
  	        
  	    }
  	    out.close();
  	}
  }
  