#include <bits/stdc++.h>
using namespace std;
const long long N = 2e5;
const long double PI = 3.14159265358979323846;
const long double eps = 1e-11;
const long long INF = 1.5e18;
const long long inf = 1e9;
const long long M = 998244353;
const string IO[2]{"NO\n", "YES\n"};
const string Io[2]{"No\n", "Yes\n"};
const string io[2]{"no\n", "yes\n"};
const string bads = "-1\n";
template <class T1, class T2>
istream& operator>>(istream& in, pair<T1, T2>& a) {
  in >> a.first >> a.second;
  return in;
}
template <class T1, class T2>
ostream& operator<<(ostream& out, pair<T1, T2>& a) {
  out << a.first << ' ' << a.second;
  return out;
}
template <class T>
istream& operator>>(istream& in, vector<T>& a) {
  for (auto& el : a) {
    in >> el;
  }
  return in;
}
template <class T>
ostream& operator<<(ostream& out, vector<T>& a) {
  for (auto& el : a) {
    out << el << ' ';
  }
  return out;
}
void solve() {
  long long n, m, k;
  cin >> n >> m >> k;
  long long ssize = n / m;
  long long bsize = (n + m - 1) / m;
  long long cnt = n % m;
  vector<vector<pair<long long, long long> > > el(k);
  long long start = 0, last_start = 0;
  for (long long i = 0; i < k; ++i) {
    start = last_start;
    for (long long j = 0; j < cnt; ++j) {
      el[i].push_back({start, (start + bsize) % n});
      start += bsize;
      start %= n;
      last_start = start;
    }
    for (long long j = 0; j < m - cnt; ++j) {
      el[i].push_back({start, (start + ssize) % n});
      start += ssize;
      start %= n;
    }
  }
  for (long long i = 0; i < k; ++i) {
    for (auto block : el[i]) {
      vector<long long> ans;
      long long i = block.first;
      bool ok = 1;
      while (i != block.second || ok) {
        ok = 0;
        ans.push_back(i++ + 1);
        i %= n;
      }
      cout << ans.size() << ' ' << ans << '\n';
    }
  }
  cout << '\n';
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  long long t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
}

### Prompt

Please formulate a CPP solution to the following problem:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const long long N = 2e5;
const long double PI = 3.14159265358979323846;
const long double eps = 1e-11;
const long long INF = 1.5e18;
const long long inf = 1e9;
const long long M = 998244353;
const string IO[2]{"NO\n", "YES\n"};
const string Io[2]{"No\n", "Yes\n"};
const string io[2]{"no\n", "yes\n"};
const string bads = "-1\n";
template <class T1, class T2>
istream& operator>>(istream& in, pair<T1, T2>& a) {
  in >> a.first >> a.second;
  return in;
}
template <class T1, class T2>
ostream& operator<<(ostream& out, pair<T1, T2>& a) {
  out << a.first << ' ' << a.second;
  return out;
}
template <class T>
istream& operator>>(istream& in, vector<T>& a) {
  for (auto& el : a) {
    in >> el;
  }
  return in;
}
template <class T>
ostream& operator<<(ostream& out, vector<T>& a) {
  for (auto& el : a) {
    out << el << ' ';
  }
  return out;
}
void solve() {
  long long n, m, k;
  cin >> n >> m >> k;
  long long ssize = n / m;
  long long bsize = (n + m - 1) / m;
  long long cnt = n % m;
  vector<vector<pair<long long, long long> > > el(k);
  long long start = 0, last_start = 0;
  for (long long i = 0; i < k; ++i) {
    start = last_start;
    for (long long j = 0; j < cnt; ++j) {
      el[i].push_back({start, (start + bsize) % n});
      start += bsize;
      start %= n;
      last_start = start;
    }
    for (long long j = 0; j < m - cnt; ++j) {
      el[i].push_back({start, (start + ssize) % n});
      start += ssize;
      start %= n;
    }
  }
  for (long long i = 0; i < k; ++i) {
    for (auto block : el[i]) {
      vector<long long> ans;
      long long i = block.first;
      bool ok = 1;
      while (i != block.second || ok) {
        ok = 0;
        ans.push_back(i++ + 1);
        i %= n;
      }
      cout << ans.size() << ' ' << ans << '\n';
    }
  }
  cout << '\n';
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  long long t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
}

```