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 MB = 1 << 20;
  struct FastIO {
    char ib[MB + 100], *p, *q;
    char ob[MB + 100], *r, stk[128];
    int tp;
    FastIO() { p = q = ib, r = ob, tp = 0; }
    ~FastIO() { fwrite(ob, 1, r - ob, stdout); }
    char read_char() {
      if (p == q) {
        p = ib, q = ib + fread(ib, 1, MB, stdin);
        if (p == q) return 0;
      }
      return *p++;
    }
    template <typename T>
    void read_int(T& x) {
      char c = read_char(), l = 0;
      for (x = 0; !isdigit(c); c = read_char()) l = c;
      for (; isdigit(c); c = read_char()) x = x * 10 - '0' + c;
      if (l == '-') x = -x;
    }
    void write_char(char c) {
      if (r - ob == MB) r = ob, fwrite(ob, 1, MB, stdout);
      *r++ = c;
    }
    template <typename T>
    void write_int(T x) {
      if (x < 0) write_char('-'), x = -x;
      do stk[++tp] = x % 10 + '0';
      while (x /= 10);
      while (tp) write_char(stk[tp--]);
    }
  } IO;
  const int N = 200010;
  int T, n, m, k;
  int main() {
    scanf("%d", &T);
    while (T--) {
      scanf("%d%d%d", &n, &m, &k);
      int k1 = n / m;
      int k2 = (n - m * k1), nex = 1, kk = 1;
      k1 = m - k2;
      for (int i = 1; i <= k; i++) {
        for (int o = 1; o <= m; o++) {
          if (o <= k1) {
            printf("%d ", n / m);
            for (int p = 1; p <= n / m; p++) {
              printf("%d ", nex++);
              if (nex > n) nex -= n;
            }
            printf("\n");
          } else {
            if (o == k1 + 1) kk = nex;
            printf("%d ", n / m + 1);
            for (int p = 1; p <= n / m + 1; p++) {
              printf("%d ", kk++);
              if (kk > n) kk -= n;
            }
            printf("\n");
          }
        }
      }
    }
    return 0;
  }
  