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 E_Gardener_and_Tree implements Runnable{
  
         public static void main(String[] args) throws Exception {
          new Thread(null, new E_Gardener_and_Tree(), "E_Gardener_and_Tree", 1<<28).start();
         }
      public void run(){
           WRITE YOUR CODE HERE!!!!
           JUST WRITE EVERYTHING HERE WHICH YOU WRITE IN MAIN!!!
         }
  
    }
  */
  /////**************************************************************************
  
  
  public class F_Let_s_Play_the_Hat_{
      public static void main(String[] args) {
          FastScanner s= new FastScanner();
          //PrintWriter out=new PrintWriter(System.out);
          //end of program
          //out.println(answer);
          //out.close();
          StringBuilder res = new StringBuilder();
          int t=s.nextInt();
          int p=0;
          while(p<t){
          long n=s.nextLong();// people
          long m=s.nextLong();//table
          long k=s.nextLong();// games
          Queue<Long> nice = new LinkedList<Long>();
          for(int i=1;i<=n;i++){
              nice.add((long)i);
          }
          long small=n/m;
          long big=small;
          if(n%m!=0){
              big++;
          }
          long count=0;// number of big tables
         
          for(long i=0;i<=m;i++){
              long hh=(i*big);
              long yo2=n-hh;
              if(yo2%small==0){
                  long a1=yo2/small;
                  if((a1+i)==m){
                      count=i;
                      break;
                  }
              }
  
          }
          for(int i=0;i<k;i++){
              HashMap<Long,Long> map = new HashMap<Long,Long>();
                for(int j=0;j<count;j++){
                    res.append(big+" ");
  
                      for(int l=0;l<big;l++){
                            long a=nice.poll();
                            res.append(a+" ");
                            map.put(a,(long)1);
                            nice.add(a);
  
                      }
                      res.append(" \n");
                }
                ArrayList<Long> well = new ArrayList<Long>();
                for(int l=1;l<=n;l++){
                    if(!map.containsKey((long)l)){
                        well.add((long)l);
                    }
                }
                long hh=well.size()/small;
                int yo1=0;
                for(int j=0;j<hh;j++){
                    res.append(small+" ");
                    for(int l=0;l<small;l++){
                        long jj=well.get(yo1);
                        res.append(jj+" ");
                        yo1++;
                    }
                    res.append(" \n");
                }
          }
          res.append(" \n");
          p++;
      }
      System.out.println(res);
  
  
  
          
      }
      static class FastScanner {
  		BufferedReader br;
  		StringTokenizer st;
   
  		public FastScanner(String s) {
  			try {
  				br = new BufferedReader(new FileReader(s));
  			} catch (FileNotFoundException e) {
  				// TODO Auto-generated catch block
  				e.printStackTrace();
  			}
  		}
   
  		public FastScanner() {
  			br = new BufferedReader(new InputStreamReader(System.in));
  		}
   
  		String nextToken() {
  			while (st == null || !st.hasMoreElements()) {
  				try {
  					st = new StringTokenizer(br.readLine());
  				} catch (IOException e) {
  					// TODO Auto-generated catch block
  					e.printStackTrace();
  				}
  			}
  			return st.nextToken();
  		}
   
  		int nextInt() {
  			return Integer.parseInt(nextToken());
  		}
   
  		long nextLong() {
  			return Long.parseLong(nextToken());
  		}
   
  		double nextDouble() {
  			return Double.parseDouble(nextToken());
  		}
  	}
  
  
      static long modpower(long x, long y, long p)
       {
        long res = 1; // Initialize result
     
        x = x % p; // Update x if it is more than or
        // equal to p
     
        if (x == 0)
          return 0; // In case x is divisible by p;
     
        while (y > 0)
        {
     
          // If y is odd, multiply x with result
          if ((y & 1) != 0)
            res = (res * x) % p;
     
          // y must be even now
          y = y >> 1; // y = y/2
          x = (x * x) % p;
        }
        return res;
      }
  
    
     // SIMPLE POWER FUNCTION=>
  
  
        static long power(long x, long y)
          {
              long res = 1;     // Initialize result
          
              while (y > 0)
              {
              
                  // If y is odd, multiply x with result
                  if ((y & 1) != 0)
                      res = res * x;
          
                  // y must be even now
                  y = y >> 1; // y = y/2
                  x = x * x;  // Change x to x^2
              }
              return res;
          }
  
  
  
  
  
  }