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: '''
      Auther: ghoshashis545 Ashis Ghosh
      College: Jalpaiguri Govt Enggineering College
  
  '''
  from os import path
  from io import BytesIO, IOBase
  import sys
  from heapq import heappush,heappop
  from functools import cmp_to_key as ctk
  from collections import deque,Counter,defaultdict as dd 
  from bisect import bisect,bisect_left,bisect_right,insort,insort_left,insort_right
  from itertools import permutations
  from datetime import datetime
  from math import ceil,sqrt,log,gcd
  def ii():return int(input())
  def si():return input().rstrip()
  def mi():return map(int,input().split())
  def li():return list(mi())
  abc='abcdefghijklmnopqrstuvwxyz'
  mod=1000000007
  #mod=998244353
  inf = float("inf")
  vow=['a','e','i','o','u']
  dx,dy=[-1,1,0,0],[0,0,1,-1]
  
  def bo(i):
      return ord(i)-ord('0')
  
  file = 1
  def ceil(a,b):
      return (a+b-1)//b
  
  
  
  
  def solve():
  
      
      for _ in range(1,ii()+1):
          n,m,k = mi()
  
          r_up = (n+m-1)//m
          r_down = n//m
  
          big_group = (n%m)*r_up
          small_group = n - big_group
          
          big_group_start = 0
          for i in range(k):
              cur = big_group_start
              for j in range(big_group):
                  if j%r_up == 0:
                      if j:
                          print()
                      print(r_up,end=" ")
                  print(cur+1,end=" ")
                  cur+=1
                  cur %= n
              if(big_group):
                  print()
              for j in range(small_group):
                  if j%r_down == 0:
                      if j:
                          print()
                      print(r_down,end=" ")
                  print(cur+1,end=" ")
                  cur+=1
                  cur %= n
              if(small_group):
                  print()
  
              big_group_start += big_group
              big_group_start %= n
          print()
          
          
  
  
  
  
  
  
  
              
  
  
  
  
  
      
          
  
  
  
  
  
  
  
  
          
  if __name__ =="__main__":
  
      if(file):
      
         if path.exists('input.txt'):
             sys.stdin=open('input.txt', 'r')
             sys.stdout=open('output.txt','w')
         else:
             input=sys.stdin.readline
      solve()
  