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 long long mod = 998244353;
  const long long sz = 2e6 + 5;
  vector<long long> fac, ifac;
  vector<int> prime;
  long long mpow(long long a, long long b = mod - 2, long long m = mod) {
    long long res = 1;
    a %= m;
    assert(b >= 0);
    for (; b; b >>= 1) {
      if (b & 1) res = res * a % m;
      a = a * a % m;
    }
    return res;
  }
  void init() {
    fac.resize(sz, 1);
    ifac.resize(sz);
    for (long long i = 2; i < sz; i++) fac[i] = (fac[i - 1] * i) % mod;
    ifac[sz - 1] = mpow(fac[sz - 1]);
    for (long long i = sz - 2; i >= 0; i--)
      ifac[i] = (ifac[i + 1] * (i + 1)) % mod;
    prime.resize(sz);
    for (int i = 2; i < sz; i++) prime[i] = i;
    for (long long i = 2; i * i < sz; i++) {
      if (prime[i] != i) continue;
      for (long long j = i * i; j < sz; j += i)
        if (prime[j] == j) prime[j] = i;
    }
  }
  long long nCr(long long n, long long r) {
    if (n <= 0 || r < 0) return 0;
    if (n < r) return 0;
    return (fac[n] * (ifac[r] * ifac[n - r] % mod)) % mod;
  }
  long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
  long long lcm(long long a, long long b) { return (a / gcd(a, b)) * b; }
  void solve() {
    long long n, m, k;
    cin >> n >> m >> k;
    long long high_tables = n % m;
    long long rem_tables = m - high_tables;
    long long high_people = (n + m - 1) / m;
    long long less_people = n / m;
    queue<long long> players;
    for (int i = 1; i <= n; i++) players.push(i);
    for (long long game = 0; game < k; game++) {
      long long table_no;
      for (table_no = 1; table_no <= high_tables; ++table_no) {
        cout << high_people << " ";
        for (int peep = 1; peep <= high_people; peep++) {
          long long indx = players.front();
          cout << indx << " ";
          players.pop();
          players.push(indx);
        }
        cout << "\n";
      }
      for (table_no = 1; table_no <= rem_tables; ++table_no) {
        cout << less_people << " ";
        for (int peep = 1; peep <= less_people; peep++) {
          long long indx = players.front();
          cout << indx << " ";
          players.pop();
          players.push(indx);
        }
        cout << "\n";
      }
      long long shift_back = high_tables * high_people;
      for (int i = 0; i < shift_back; i++) {
        long long indx = players.front();
        players.pop();
        players.push(indx);
      }
    }
    cout << "\n";
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    init();
    int t = 1;
    cin >> t;
    for (int i = 1; i <= t; i++) {
      solve();
    }
    return 0;
  }
  