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;
  const int INF = 0x3f3f3f3f;
  const long long LINF = 0x3f3f3f3f3f3f3f3fll;
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    int t;
    cin >> t;
    while (t--) {
      int n, m, k;
      cin >> n >> m >> k;
      if ((n % m) == 0) {
        for (int i = 0; i < k; i++) {
          int at = 0;
          for (int j = 0; j < m; j++) {
            cout << n / m << " ";
            for (int x = 0; x < n / m; x++) {
              cout << at + 1 << " ";
              at = at + 1;
            }
            cout << '\n';
          }
        }
        cout << '\n';
        continue;
      }
      vector<int> w(m);
      {
        int at = 0;
        for (int i = 0; i < n; i++) {
          w[at]++;
          at = (at + 1) % m;
        }
      }
      int mi = *min_element(w.begin(), w.end());
      int peq = 0;
      for (int i : w)
        if (i == mi) peq++;
      int last = 0;
      for (int i = 0; i < k; i++) {
        int ini = last;
        int first = ini;
        for (int j = 0; j < peq; j++) {
          cout << n / m << " ";
          for (int x = 0; x < n / m; x++) {
            cout << ini + 1 << " ";
            ini = (ini + 1) % n;
          }
          cout << '\n';
        }
        last = ini;
        while (first != ini) {
          cout << (n + m - 1) / m << " ";
          for (int x = 0; x < (n + m - 1) / m; x++) {
            cout << ini + 1 << " ";
            ini = (ini + 1) % n;
          }
          cout << '\n';
        }
      }
      cout << '\n';
    }
    exit(0);
  }
  