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 _762 {
      public static void main(String[] args) {
          MyScanner sc = new MyScanner();
          PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
          int t = sc.nextInt();
          while (t-- > 0) {
              int n = sc.nextInt();
              int m = sc.nextInt();
              int k = sc.nextInt();
              int extra = n % m;
              int small = n / m;
              int total = extra * small + extra;
              int big = small + 1;
              if (total <= n / 2) {
                  int start = 0;
                  for (int turn = 0; turn < k; turn++) {
                      Set<Integer> set = new HashSet<>();
                      for (int i = 1; i <= n; i++) set.add(i);
                      for (int i = 0; i < extra; i++) {
                          out.println();
                          out.print(big + " ");
                          for (int j = (i * big + start); j < ((i + 1) * big + start); j++) {
                              int x = j % n;
                              out.print((x + 1) + " ");
                              set.remove(x + 1);
                          }
                      }
                      int count = 0;
                      for (Integer i: set) {
                          if (count % small == 0) {
                              out.println();
                              out.print(small + " ");
                          }
                          out.print(i + " ");
                          ++count;
                      }
                      start += total;
                      out.println();
                  }
              } else {
                  total = n - total;
                  extra = m - extra;
                  int start = 0;
                  for (int turn = 0; turn < k; turn++) {
                      Set<Integer> set = new HashSet<>();
                      for (int i = 1; i <= n; i++) set.add(i);
                      for (int i = 0; i < extra; i++) {
                          out.println();
                          out.print(small + " ");
                          for (int j = (i * small + start); j < ((i + 1) * small + start); j++) {
                              int x = j % n;
                              out.print((x + 1) + " ");
                              set.remove(x + 1);
                          }
                      }
                      int count = 0;
                      for (Integer i: set) {
                          if (count % big == 0) {
                              out.println();
                              out.print(big + " ");
                          }
                          out.print(i + " ");
                          ++count;
                      }
                      start = (start + total) % n;
                      out.println();
                  }
              }
          }
          out.close();
      }
  
  
      static void sort(int[] a) {
          ArrayList<Integer> q = new ArrayList<>();
          for (int i : a) q.add(i);
          Collections.sort(q);
          for (int i = 0; i < a.length; i++) a[i] = q.get(i);
      }
  
      static void sort(long[] a) {
          ArrayList<Long> q = new ArrayList<>();
          for (long i : a) q.add(i);
          Collections.sort(q);
          for (int i = 0; i < a.length; i++) a[i] = q.get(i);
      }
  
  
      //-----------MyScanner class for faster input----------
      public static class MyScanner {
          BufferedReader br;
          StringTokenizer st;
  
          public MyScanner() {
              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;
          }
  
  
      }
  
  }