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 sys
  input = sys.stdin.readline
  
  t = int(input())
  ans = []
  for _ in range(t):
      n, m, k = map(int, input().split())
      if n % m == 0:
          c = n // m
          ans0 = []
          for _ in range(k):
              for i in range(m):
                  ans0.append(" ".join(map(str, [c] + [j + i * c + 1 for j in range(c)])))
          ans.append("\n".join(ans0))
          continue
      x = [n // m + min(n % m, 1)] * m
      s = sum(x)
      u = 0
      for i in range(s - n):
          x[i] -= 1
          u += x[i]
      c = u * k
      y = [set() for _ in range(k)]
      z = [c // n + min(c % n, 1)] * n
      s = sum(z)
      for i in range(s - c):
          z[i] -= 1
      j = 0
      for i in range(n):
          for _ in range(z[i]):
              y[j].add(i + 1)
              j += 1
              j %= k
      ans0 = []
      for i in range(k):
          a = y[i]
          b = set()
          for j in range(1, n + 1):
              if not j in a:
                  b.add(j)
          a = list(a)
          b = list(b)
          for j in x:
              ans00 = [j]
              if j == x[0]:
                  for _ in range(j):
                      ans00.append(a.pop())
              else:
                  for _ in range(j):
                      ans00.append(b.pop())
              ans0.append(" ".join(map(str, ans00)))
      ans.append("\n".join(ans0))
  sys.stdout.write("\n\n".join(ans))