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 = pow(10, 9) + 7;
  long long fac[1001];
  template <typename T, typename U>
  static inline void amin(T &x, U y) {
    if (y < x) x = y;
  }
  template <typename T, typename U>
  static inline void amax(T &x, U y) {
    if (x < y) x = y;
  }
  void SieveOfEratosthenes(long long n) {
    bool prime[n + 1];
    memset(prime, true, sizeof(prime));
    for (long long p = 2; p * p <= n; p++) {
      if (prime[p] == true) {
        for (long long i = p * p; i <= n; i += p) prime[i] = false;
      }
    }
    for (long long p = 2; p <= n; p++)
      if (prime[p]) {
      }
  }
  long long power(long long x, long long y, long long p) {
    long long res = 1;
    x = x % p;
    while (y > 0) {
      if (y & 1) res = (res * x) % p;
      y = y >> 1;
      x = (x * x) % p;
    }
    return res;
  }
  long long modInverse(long long n, long long p) { return power(n, p - 2, p); }
  long long nCrModPFermat(long long n, long long r, long long p) {
    if (n < r) return 0;
    if (r == 0) return 1;
    return (fac[n] * modInverse(fac[r], p) % p * modInverse(fac[n - r], p) % p) %
           p;
  }
  void dfs(long long v, vector<vector<long long>> &adj,
           vector<long long> &visited) {
    for (long long u : adj[v]) {
      if (!visited[u]) {
        visited[u] = 1;
      }
    }
  }
  long long parent[300002], sz[300002];
  long long find_set(long long v) {
    if (v == parent[v]) return v;
    return parent[v] = find_set(parent[v]);
  }
  void make_set(long long v) {
    parent[v] = v;
    sz[v] = 1;
  }
  void union_sets(long long a, long long b) {
    a = find_set(a);
    b = find_set(b);
    if (a != b) {
      if (sz[a] < sz[b]) swap(a, b);
      parent[b] = a;
      sz[a] += sz[b];
    }
  }
  class segTree {
   public:
    long long len;
    vector<long long> a;
    vector<pair<long long, long long>> t;
    segTree(vector<long long> &val) {
      len = val.size();
      a.resize(len);
      long long i;
      a = val;
      if ((len & (len - 1)) != 0) {
        long long np = ceil(log2(len));
        long long pval = pow(2, np);
        long long diff = pval - len;
        len = pval;
        for (i = 0; i < diff; i++) a.push_back(pow(10, 10));
      }
      t.resize(4 * len + 1);
    }
    void build(long long idx, long long tl, long long tr) {
      if (tl == tr) {
        t[idx] = {a[tl], 1};
        return;
      }
      long long tm = (tl + tr) / 2;
      build(idx * 2, tl, tm);
      build(idx * 2 + 1, tm + 1, tr);
      long long count = 0;
      long long mn = min(t[2 * idx].first, t[2 * idx + 1].first);
      if (mn == t[2 * idx].first) count += t[2 * idx].second;
      if (mn == t[2 * idx + 1].first) count += t[2 * idx + 1].second;
      t[idx] = {mn, count};
    }
    pair<long long, long long> query(long long idx, long long tl, long long tr,
                                     long long l, long long r) {
      if (l > r) return {pow(10, 10), 1};
      if (l == tl && r == tr) {
        return t[idx];
      }
      long long tm = (tl + tr) / 2;
      pair<long long, long long> res1 = query(idx * 2, tl, tm, l, min(r, tm));
      pair<long long, long long> res2 =
          query(idx * 2 + 1, tm + 1, tr, max(l, tm + 1), r);
      if (res1.first < res2.first) {
        return res1;
      } else if (res2.first < res1.first) {
        return res2;
      } else {
        return {res1.first, res1.second + res2.second};
      }
    }
    void update(long long idx, long long tl, long long tr, long long pos,
                long long val) {
      if (tl == tr) {
        t[idx] = {val, 1};
        return;
      }
      long long tm = (tl + tr) / 2;
      if (pos <= tm)
        update(2 * idx, tl, tm, pos, val);
      else
        update(2 * idx + 1, tm + 1, tr, pos, val);
      long long count = 0;
      long long mn = min(t[2 * idx].first, t[2 * idx + 1].first);
      if (mn == t[2 * idx].first) count += t[2 * idx].second;
      if (mn == t[2 * idx + 1].first) count += t[2 * idx + 1].second;
      t[idx] = {mn, count};
    }
  };
  void solve() {
    long long i, j, n, m, k;
    cin >> n >> m >> k;
    if (n % m == 0) {
      for (i = 1; i <= k; i++) {
        for (j = 1; j <= m; j++) {
          long long z = n / m;
          cout << z << " ";
          long long s = z * (j - 1) + 1;
          long long e = z * j;
          for (int r = s; r <= e; r++) {
            cout << r << " ";
          }
          cout << '\n';
        }
      }
      return;
    }
    long long big = n % m, sm = m - big;
    long long c = 1;
    for (i = 1; i <= k; i++) {
      long long z = n / m;
      for (j = 1; j <= big; j++) {
        cout << z + 1 << " ";
        for (long long r = 1; r <= z + 1; r++) {
          cout << c << " ";
          c++;
          if (c > n) c = 1;
        }
        cout << '\n';
      }
      long long s = c;
      for (j = 1; j <= sm; j++) {
        cout << z << " ";
        for (long long r = 1; r <= z; r++) {
          cout << s << " ";
          s++;
          if (s > n) s = 1;
        }
        cout << '\n';
      }
    }
    return;
  }
  int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    long long test = 2;
    switch (test) {
      case 1:
        solve();
        break;
      case 2: {
        long long t;
        cin >> t;
        while (t--) {
          solve();
        }
      }; break;
    }
  }
  