#include <bits/stdc++.h>
using namespace std;
void printvec(vector<long long int> &vec) {
  for (long long int i = 0; i < vec.size(); i++) cout << vec[i] << " ";
  cout << '\n';
}
const long long int INF_ = 1e12;
long long int n, m, k;
multiset<pair<long long int, long long int> > mt;
void printmt() {
  for (auto &m : mt) {
    cout << "(" << m.first << "," << m.second << ") ";
  }
  cout << '\n';
}
void printst(set<long long int> &st) {
  for (auto &s : st) {
    cout << s << " ";
  }
  cout << '\n';
}
void reset() {
  mt.clear();
  for (long long int i = 1; i <= n; i++) {
    mt.insert(make_pair(0, i));
  }
}
void make_gps(set<long long int> &st, long long int sz) {
  long long int i = 0;
  vector<long long int> temp_vec;
  for (auto &s : st) {
    if (i == 0) {
      cout << sz << " ";
    }
    temp_vec.push_back(s);
    i++;
    if (i == sz) {
      printvec(temp_vec);
      temp_vec.clear();
      i = 0;
    }
  }
}
void solve() {
  cin >> n >> m >> k;
  reset();
  long long int nb = (n % m) * ((n + m - 1) / m);
  long long int sz_big = (n + m - 1) / m;
  long long int sz_small = n / m;
  for (long long int i = 0; i < k; i++) {
    set<long long int> bigs, smalls;
    for (long long int j = 0; j < nb; j++) {
      pair<long long int, long long int> temp = *(mt.begin());
      bigs.insert(temp.second);
      mt.erase(mt.find(temp));
      mt.insert(make_pair(temp.first + 1, temp.second));
    }
    for (long long int j = 1; j <= n; j++) {
      if (bigs.find(j) == bigs.end()) {
        smalls.insert(j);
      }
    }
    make_gps(bigs, sz_big);
    make_gps(smalls, sz_small);
  }
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int t;
  cin >> t;
  while (t--) {
    solve();
  }
}

### Prompt

Create a solution in Cpp for 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;
void printvec(vector<long long int> &vec) {
  for (long long int i = 0; i < vec.size(); i++) cout << vec[i] << " ";
  cout << '\n';
}
const long long int INF_ = 1e12;
long long int n, m, k;
multiset<pair<long long int, long long int> > mt;
void printmt() {
  for (auto &m : mt) {
    cout << "(" << m.first << "," << m.second << ") ";
  }
  cout << '\n';
}
void printst(set<long long int> &st) {
  for (auto &s : st) {
    cout << s << " ";
  }
  cout << '\n';
}
void reset() {
  mt.clear();
  for (long long int i = 1; i <= n; i++) {
    mt.insert(make_pair(0, i));
  }
}
void make_gps(set<long long int> &st, long long int sz) {
  long long int i = 0;
  vector<long long int> temp_vec;
  for (auto &s : st) {
    if (i == 0) {
      cout << sz << " ";
    }
    temp_vec.push_back(s);
    i++;
    if (i == sz) {
      printvec(temp_vec);
      temp_vec.clear();
      i = 0;
    }
  }
}
void solve() {
  cin >> n >> m >> k;
  reset();
  long long int nb = (n % m) * ((n + m - 1) / m);
  long long int sz_big = (n + m - 1) / m;
  long long int sz_small = n / m;
  for (long long int i = 0; i < k; i++) {
    set<long long int> bigs, smalls;
    for (long long int j = 0; j < nb; j++) {
      pair<long long int, long long int> temp = *(mt.begin());
      bigs.insert(temp.second);
      mt.erase(mt.find(temp));
      mt.insert(make_pair(temp.first + 1, temp.second));
    }
    for (long long int j = 1; j <= n; j++) {
      if (bigs.find(j) == bigs.end()) {
        smalls.insert(j);
      }
    }
    make_gps(bigs, sz_big);
    make_gps(smalls, sz_small);
  }
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int t;
  cin >> t;
  while (t--) {
    solve();
  }
}

```