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.*;
   
  public class Solution extends PrintWriter {
      
      
      void solve() {
          int t = sc.nextInt();
          for(int i = 1; i <= t; i++) {
              test_case();
          }
      }
      
      void test_case() {
          int n = sc.nextInt();
          int m = sc.nextInt();
          int k = sc.nextInt();
          PriorityQueue<int[]> pq = new PriorityQueue<int[]>(Comparator.comparing(arr -> arr[0]));
          PriorityQueue<int[]> next = new PriorityQueue<int[]>(Comparator.comparing(arr -> arr[0]));
          int f = n/m;
          int c = (n+m-1)/m;
          for(int i = 1; i <= n; i++) pq.add(new int[] {0, i});
          for(int it = 0; it < k; it++) {
              for(int t = 0; t < m; t++) {
                  int len = f;
                  if(t < n%m) {
                      len = c;
                  }
                  print(len + " ");
                  for(int i = 0; i < len; i++) {
                      int[] cur = pq.remove();
                      print(cur[1] + " ");
                      if(t < n%m) cur[0]++;
                      next.add(cur);
                  }
                  println();
              }
              PriorityQueue<int[]> temp = pq;
              pq = next;
              next = temp;
          }
          println();
      }
      
          
  //  Main() throws FileNotFoundException { super(new File("output.txt")); }
  //  InputReader sc = new InputReader(new FileInputStream("test_input.txt"));
    Solution() { super(System.out); }
    InputReader sc = new InputReader(System.in);
    static class InputReader {
        InputReader(InputStream in) { this.in = in; } InputStream in;
        
        private byte[] buf = new byte[16384];
        private int    curChar;
        private int    numChars;
        
   
        public int read() {
            if (numChars == -1)
                throw new InputMismatchException();
            if (curChar >= numChars) {
                curChar = 0;
                try {
                    numChars = in.read(buf);
                } catch (IOException e) {
                    throw new InputMismatchException();
                }
                if (numChars <= 0)
                    return -1;
            }
            return buf[curChar++];
        }
   /*
        private String nextLine() {
            int c = read();
            while (isSpaceChar(c))
                c = read();
            StringBuilder res = new StringBuilder();
            do {
                res.appendCodePoint(c);
                c = read();
            } while (!isEndOfLine(c));
            return res.toString();
        }*/
   
        public String nextString() {
            int c = read();
            while (isSpaceChar(c))
                c = read();
            StringBuilder res = new StringBuilder();
            do {
                res.appendCodePoint(c);
                c = read();
            } while (!isSpaceChar(c));
            return res.toString();
        }
   
        public long nextLong() {
            int c = read();
            while (isSpaceChar(c))
                c = read();
            int sgn = 1;
            if (c == '-') {
                sgn = -1;
                c = read();
            }
            long res = 0;
            do {
                if (c < '0' || c > '9')
                    throw new InputMismatchException();
                res *= 10;
                res += c - '0';
                c = read();
            } while (!isSpaceChar(c));
            return res * sgn;
        }
   
        public int nextInt() {
            int c = read();
            while (isSpaceChar(c))
                c = read();
            int sgn = 1;
            if (c == '-') {
                sgn = -1;
                c = read();
            }
            int res = 0;
            do {
                if (c < '0' || c > '9')
                    throw new InputMismatchException();
                res *= 10;
                res += c - '0';
                c = read();
            } while (!isSpaceChar(c));
            return res * sgn;
        }
   
        private boolean isSpaceChar(int c) {
            return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
        }
   
        private boolean isEndOfLine(int c) {
            return c == '\n' || c == '\r' || c == -1;
        }
    }
   
      public static void main(String[] $) {
          new Thread(null, new Runnable() {
              public void run() {
                  long start = System.nanoTime();
                  try {Solution solution = new Solution(); solution.solve(); solution.flush();} 
                  catch (Exception e) {e.printStackTrace(); System.exit(1);}
                  System.err.println((System.nanoTime()-start)/1E9);
              }
          }, "1", 1 << 27).start();
   
      }
  }