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;
  long long binpow(long long base, long long ex,
                   long long mod = (long long)1000000007) {
    long long ans = 1;
    base = base % mod;
    if (base == 0) {
      return 0;
    }
    while (ex > 0) {
      if (ex % 2 == 1) {
        ans = (ans * base) % mod;
      }
      base = (base * base) % mod;
      ex = ex / 2;
    }
    return ans;
  }
  void read(vector<long long> &arr, long long n) {
    if (arr.size() != n) {
      arr.assign(n, 0);
    }
    for (int i = 0; i < n; i++) cin >> arr[i];
  }
  inline long long min(long long a, long long b) {
    if (a > b) return b;
    return a;
  }
  inline long long max(long long a, long long b) {
    if (a > b) return a;
    return b;
  }
  inline long long dif(long long a, long long b) {
    if (a > b) return a - b;
    return b - a;
  }
  long long gcd(long long a, long long b) {
    if (b == 0) return a;
    return gcd(b, a % b);
  }
  long long lcm(long long a, long long b) {
    long long k = gcd(a, b);
    return (a * b) / k;
  }
  void solve() {
    long long n, m, k;
    cin >> n >> m >> k;
    vector<vector<vector<long long>>> ans(k, vector<vector<long long>>(2));
    long long quo = n / m;
    long long rem = n % m;
    long long constant = k * ((n % m) * (n / m + 1));
    long long x = ((constant) + (n - (constant) % n)) / n - 1;
    long long a = (constant % n);
    vector<long long> arr(n, x);
    for (long long i = 0; (i) < long(a); (i)++) {
      arr[i]++;
    }
    long long last = 0;
    for (int curr = 0; curr <= 4 * n; curr++) {
      int i = curr % n;
      long long new_last = last;
      for (int game = 0; game < k; game++) {
        if (arr[i] and ans[(game + last) % k][0].size() < (quo + 1) * rem) {
          arr[i]--;
          new_last = (game + last) % k;
          ans[(game + last) % k][0].push_back(i + 1);
        } else if (ans[(game + last) % k][1].size() < (quo) * (n - rem)) {
          ans[(game + last) % k][1].push_back(i + 1);
        }
      }
      last = (new_last + 1) % k;
    }
    for (int game = 0; game < k; game++) {
      int up = 0;
      for (int count = 0; count < rem; count++) {
        cout << quo + 1 << " ";
        for (int i = 0; i <= quo; i++) {
          cout << ans[game][0][up] << " ";
          up++;
        }
        cout << endl;
      }
      int down = 0;
      for (int count = 0; count < (m - rem); count++) {
        cout << quo << " ";
        for (int i = 0; i < quo; i++) {
          cout << ans[game][1][down] << " ";
          down++;
        }
        cout << endl;
      }
    }
    cout << endl;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout << fixed << showpoint;
    cout << setprecision(12);
    long long test_m = 1;
    cin >> test_m;
    while (test_m--) {
      solve();
    }
  }
  