#include <bits/stdc++.h>
using namespace std;
const long long Max = 1e5 + 1;
void nos() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
}
long long n, m, k;
void read() { cin >> n >> m >> k; }
void solve() {
  long long highcnt = n - (m * (n / m));
  highcnt *= n / m + 1;
  long long plusonetotal = highcnt * k;
  long long small, smallcnt, big, bigcnt;
  small = plusonetotal / n;
  big = small + 1;
  bigcnt = plusonetotal - small * n;
  smallcnt = n - bigcnt;
  vector<pair<long long, long long> > mp;
  long long i;
  for (i = 1; i <= n; i++) {
    if (bigcnt) {
      bigcnt--;
      mp.push_back({i, big});
    } else
      mp.push_back({i, small});
  }
  while (k--) {
    long long attable = n / m + 1;
    std::vector<long long> mult;
    long long luat = 0;
    vector<pair<long long, long long> > newmp;
    for (auto &it : mp) {
      if (luat < highcnt) {
        luat++;
        mult.push_back(it.first);
        it.second--;
      }
      if (it.second != 0) newmp.push_back(it);
    }
    sort(newmp.begin(), newmp.end(),
         [](pair<long long, long long> p1, pair<long long, long long> p2) {
           return p1.second > p2.second;
         });
    mp = newmp;
    long long bigline = mult.size() / attable;
    long long it = 0;
    vector<long long> fq(n + 1);
    while (bigline--) {
      cout << attable << ' ';
      long long end = it + attable;
      for (; it < end; it++) {
        fq[mult[it]] = 1;
        cout << mult[it] << ' ';
      }
      cout << '\n';
    }
    vector<long long> less;
    for (i = 1; i <= n; i++)
      if (!fq[i]) less.push_back(i);
    attable--;
    long long smallline = less.size() / attable;
    it = 0;
    while (smallline--) {
      cout << attable << ' ';
      long long end = it + attable;
      for (; it < end; it++) {
        cout << less[it] << ' ';
      }
      cout << '\n';
    }
  }
  cout << '\n';
}
void restart() {}
int32_t main() {
  nos();
  long long teste;
  cin >> teste;
  while (teste--) {
    read();
    solve();
    restart();
  }
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 Max = 1e5 + 1;
void nos() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
}
long long n, m, k;
void read() { cin >> n >> m >> k; }
void solve() {
  long long highcnt = n - (m * (n / m));
  highcnt *= n / m + 1;
  long long plusonetotal = highcnt * k;
  long long small, smallcnt, big, bigcnt;
  small = plusonetotal / n;
  big = small + 1;
  bigcnt = plusonetotal - small * n;
  smallcnt = n - bigcnt;
  vector<pair<long long, long long> > mp;
  long long i;
  for (i = 1; i <= n; i++) {
    if (bigcnt) {
      bigcnt--;
      mp.push_back({i, big});
    } else
      mp.push_back({i, small});
  }
  while (k--) {
    long long attable = n / m + 1;
    std::vector<long long> mult;
    long long luat = 0;
    vector<pair<long long, long long> > newmp;
    for (auto &it : mp) {
      if (luat < highcnt) {
        luat++;
        mult.push_back(it.first);
        it.second--;
      }
      if (it.second != 0) newmp.push_back(it);
    }
    sort(newmp.begin(), newmp.end(),
         [](pair<long long, long long> p1, pair<long long, long long> p2) {
           return p1.second > p2.second;
         });
    mp = newmp;
    long long bigline = mult.size() / attable;
    long long it = 0;
    vector<long long> fq(n + 1);
    while (bigline--) {
      cout << attable << ' ';
      long long end = it + attable;
      for (; it < end; it++) {
        fq[mult[it]] = 1;
        cout << mult[it] << ' ';
      }
      cout << '\n';
    }
    vector<long long> less;
    for (i = 1; i <= n; i++)
      if (!fq[i]) less.push_back(i);
    attable--;
    long long smallline = less.size() / attable;
    it = 0;
    while (smallline--) {
      cout << attable << ' ';
      long long end = it + attable;
      for (; it < end; it++) {
        cout << less[it] << ' ';
      }
      cout << '\n';
    }
  }
  cout << '\n';
}
void restart() {}
int32_t main() {
  nos();
  long long teste;
  cin >> teste;
  while (teste--) {
    read();
    solve();
    restart();
  }
  return 0;
}

```