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.*;
  import java.util.*;
  import java.math.*;
  
  public class A6 {
  
      void solve() {
          int n = cint();
          int m = cint();
          int k = cint();
          
          int[] a = new int[m];
          boolean[] b = new boolean[m];
          Arrays.fill(a, n / m);
          int diff = n % m;
          for (int i = 0; i < diff; i++) {
              a[i]++;
              b[i] = true;
          }
          
          TreeSet<P> treeSet = new TreeSet<>(Comparator.comparingInt((P o) -> o.b).thenComparingInt((P o) -> o.index));
          for (int i = 1; i <= n; i++) {
              treeSet.add(new P(i));
          }
  
          for (int f = 0; f < k; f++) {
              StringBuilder buf = new StringBuilder();
              List<P> line = new ArrayList<>(treeSet);
              Iterator<P> iter = line.iterator();
              for (int i = 0; i < m; i++) {
                  buf.append(a[i]);
                  for (int j = 0; j < a[i]; j++) {
                      P p = iter.next();
                      buf.append(' ').append(p.index);
                      if (b[i]) {
                          treeSet.remove(p);
                          p.b++;
                          treeSet.add(p);
                      }
                  }
                  buf.append('\n');
              }
              cout(buf.toString());
          }
      }
  
      class P {
          int index;
          int b;
  
          public P(int index) {
              this.index = index;
          }
      }
  
      public static void main(String... args) {
          int t = in.nextInt();
          for (int test = 0; test < t; test++) {
              new A6().solve();
          }
          out.flush();
      }
  
      static final QuickReader in = new QuickReader(System.in);
      static final PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out, 131_072));
  
      static int cint() {
          return in.nextInt();
      }
  
      static long clong() {
          return in.nextLong();
      }
  
      static String cstr() {
          return in.nextLine();
      }
  
      static int[] carr_int(int n) {
          return in.nextInts(n);
      }
  
      static long[] carr_long(int n) {
          return in.nextLongs(n);
      }
  
      static void cout(int... a) {
          StringBuilder buf = new StringBuilder();
          for (int i = 0; i < a.length; i++) {
              if (i != 0) {
                  buf.append(' ');
              }
              buf.append(a[i]);
          }
          out.println(buf);
      }
  
      static void cout(long... a) {
          StringBuilder buf = new StringBuilder();
          for (int i = 0; i < a.length; i++) {
              if (i != 0) {
                  buf.append(' ');
              }
              buf.append(a[i]);
          }
          out.println(buf);
      }
  
      static void cout(Object o) {
          out.println(o);
      }
  
      static class QuickReader {
          BufferedReader in;
          StringTokenizer token;
  
          public QuickReader(InputStream ins) {
              in = new BufferedReader(new InputStreamReader(ins));
              token = new StringTokenizer("");
          }
  
          public boolean hasNext() {
              while (!token.hasMoreTokens()) {
                  try {
                      String s = in.readLine();
                      if (s == null) {
                          return false;
                      }
                      token = new StringTokenizer(s);
                  } catch (IOException e) {
                      throw new InputMismatchException();
                  }
              }
              return true;
          }
  
          public String next() {
              hasNext();
              return token.nextToken();
          }
  
          public String nextLine() {
              try {
                  String s = in.readLine();
                  token = new StringTokenizer("");
                  return s;
              } catch (IOException e) {
                  throw new InputMismatchException();
              }
          }
  
          public int nextInt() {
              return Integer.parseInt(next());
          }
  
          public int[] nextInts(int n) {
              int[] res = new int[n];
              for (int i = 0; i < n; i++) {
                  res[i] = nextInt();
              }
              return res;
          }
  
          public long nextLong() {
              return Long.parseLong(next());
          }
  
          public long[] nextLongs(int n) {
              long[] res = new long[n];
              for (int i = 0; i < n; i++) {
                  res[i] = nextLong();
              }
              return res;
          }
      }
  }
  