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 Solution {
      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;
          }
      }
      static void sort(int a[]){ // int -> long
          ArrayList<Integer> arr=new ArrayList<>(); // Integer -> Long
          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);
          
      }
      private static long gcd(long a, long b){
          if(b==0)return a;
          return gcd(b,a%b);
      }
      private static long pow(long x,long y){
          if(y==0)return 1;
          long temp = pow(x, y/2);
          if(y%2==1){
              return x*temp*temp;
          }
          else{
              return temp*temp;
          }
      }
      static int log(long n){
          int res = 0;
          while(n>0){
              res++;
              n/=2;
          }
          return res;
      }
  
      static int mod = (int)1e9+7;
      static PrintWriter out;
      static FastReader sc ;
      public static void main(String[] args) throws IOException {
          sc = new FastReader();
          out = new PrintWriter(System.out);
          // primes();
          // ================================ //
          int test = sc.nextInt();
          while (test-- > 0) {
              int n = sc.nextInt();
              int m = sc.nextInt();
              int k = sc.nextInt();
              solver(n, m, k);
              out.println( );
          }
          // ================================ //
          // int n = sc.nextInt();
          // solver();
          // ================================ //
          out.flush();
      }
  
      public static void solver(int n, int m, int k) {
          Deque<Integer> q = new LinkedList<>();
          for(int i=1;i<=n;i++)q.addLast(i);
          int len = (n+m-1)/m;
                  // System.out.println("__"+    (n-(len*m-n)));
          for(int j=0;j<k;j++){
              int cnt =0;
              Stack<Integer> st = new Stack<>();
              for(int l=0;l<m;l++){
                  int curl = 1;
                  if(m-(len*m-n)>l){
                      print(len);
                      curl = len;
                  }
                  else{
                      print(len-1);
                      curl = len-1;
                  }
                  for(int i=0;i<curl;i++){
                      int temp = q.removeFirst();
                      cnt++;
                      print(temp); 
                      if(cnt>Math.min(len*(m-(len*m-n)), (len-1)*(len*m-n))){
                          st.push(temp);
                      }
                      else{
                          q.addLast(temp);
                      }
                  }
                  out.println( );
              }
              while(!st.isEmpty()){
                  q.addFirst(st.pop());
              }
          }
      }
      static void print(int n){out.print(n+" ");}
  }
  