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: #include <bits/stdc++.h>
  using namespace std;
  void err() { cout << "\033[39;0m" << endl; }
  template <class T, class... Ts>
  void err(const T& arg, const Ts&... args) {
    cout << arg << " ";
    err(args...);
  }
  const int N = 200010;
  int T, n, m, k, v[N];
  void solve() {
    if (n % m == 0) {
      for (int p = int(1); p <= int(k); p++) {
        for (int j = int(1); j <= int(m); j++) {
          printf("%d ", n / m);
          for (int i = (j - 1) * (n / m) + 1; i <= j * (n / m); i++) {
            printf("%d ", i);
          }
          puts("");
        }
      }
      return;
    }
    int a = n % m, b = m - a;
    int up = n / m + 1;
    int down = n / m;
    int i = 1, num = a * up;
    for (int p = int(1); p <= int(k); p++) {
      for (int l = int(1); l <= int(n); l++) v[l] = 0;
      for (int j = int(1); j <= int(a); j++) {
        printf("%d ", up);
        for (int l = 1; l <= up; l++) {
          printf("%d ", i);
          v[i] = 1;
          i = i % n + 1;
        }
        puts("");
      }
      int cnt = 0;
      for (int l = int(1); l <= int(n); l++) {
        if (v[l] == 0) {
          if (cnt % down == 0) {
            if (cnt > 0) puts("");
            printf("%d ", down);
          }
          printf("%d ", l);
          cnt++;
        }
      }
      puts("");
    }
  }
  int main() {
    cin >> T;
    while (T--) {
      scanf("%d%d%d", &n, &m, &k);
      solve();
    }
  }
  