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;
  inline void wt(long long x) { cout << x; }
  inline void wtl(long long x) { cout << x << '\n'; }
  inline void wtb(long long x) { cout << x << ' '; }
  inline void wt(char x) { cout << x; }
  inline void wtl(char x) { cout << x << '\n'; }
  inline void wtb(char x) { cout << x << ' '; }
  inline void wt(string x) { cout << x; }
  inline void wtl(string x) { cout << x << '\n'; }
  inline void wtb(string x) { cout << x << ' '; }
  template <typename T>
  bool ckmax(T &x, T y) {
    return x < y ? x = y, true : false;
  }
  template <typename T>
  bool ckmin(T &x, T y) {
    return x > y ? x = y, true : false;
  }
  long long qmi(long long a, long long k, long long p) {
    long long res = 1;
    while (k) {
      if (k & 1) res = (long long)res * a % p;
      a = (long long)a * a % p;
      k >>= 1;
    }
    return res;
  }
  long long qpow(long long a, long long b) {
    long long res = 1;
    while (b) {
      if (b & 1) res *= a;
      b >>= 1;
      a *= a;
    }
    return res;
  }
  long long mo(long long x, long long p) { return x = ((x % p) + p) % p; }
  long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
  const long long maxn = 1e6 + 7;
  const long long mod = 1e9 + 7;
  long long dx[] = {0, 0, 1, -1}, dy[] = {1, -1, 0, 0};
  long long fact[maxn], infact[maxn];
  void jie_init(long long n) {
    fact[0] = 1;
    for (long long i = (1); i <= (n); i++) fact[i] = fact[i - 1] * i % mod;
    infact[n] = qmi(fact[n], mod - 2, mod);
    for (long long i = (n - 1); i >= (0); i--)
      infact[i] = infact[i + 1] * (i + 1) % mod;
  }
  long long C(long long n, long long m) {
    return fact[n] * infact[m] % mod * infact[n - m] % mod;
  }
  long long T = 1, N, M, K;
  long long A[maxn], B[maxn], dp[maxn], f[maxn];
  void solve() {
    cin >> N >> M >> K;
    long long blo = N / M;
    long long up = N % M;
    long long down = M - up;
    long long now = 1;
    for (long long i = (1); i <= (K); i++) {
      for (long long j = (1); j <= (down); j++) {
        wtb(blo);
        for (long long k = (1); k <= (blo); k++) {
          wtb(now);
          now++;
          if (now > N) now = 1;
        }
        cout << '\n';
      }
      long long temp = now;
      for (long long j = (1); j <= (up); j++) {
        wtb(blo + 1);
        for (long long k = (1); k <= (blo + 1); k++) {
          wtb(temp);
          temp++;
          if (temp > N) temp = 1;
        }
        cout << '\n';
      }
    }
  }
  signed main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin >> T;
    while (T--) solve();
    return (0 - 0);
  }
  