#include <bits/stdc++.h>
const long long N = 1e6 + 10;
const long long INF = 1e18;
const long long inf = -1e18;
const long long MOD = 998244353;
using namespace std;
long long n, m, k;
long long inv(long long a, long long m) {
  long long temp = m, q, t, u = 0, v = 1;
  if (m == 1) return 0;
  while (a > 1) {
    q = a / m;
    t = m;
    m = a % m;
    a = t;
    t = u;
    u = v - q * u;
    v = t;
  }
  if (v < 0) v += temp;
  return v;
}
int main() {
  ios_base::sync_with_stdio(0);
  int o = 1;
  cin >> o;
  while (o--) {
    cin >> n >> m >> k;
    deque<long long> a;
    for (int i = 0; i < n; i++) {
      a.push_back(i + 1);
    }
    for (int i = 1; i <= k; i++) {
      long long cnt = 1;
      for (int j = 1; j <= m; j++) {
        if (j <= n % m) {
          cout << n / m + 1 << " ";
          for (int j = cnt; j <= cnt + n / m + 1 - 1; j++) {
            cout << a[j - 1] << " ";
          }
          cnt += n / m + 1;
        } else {
          cout << n / m << " ";
          for (int j = cnt; j <= cnt + n / m - 1; j++) {
            cout << a[j - 1] << " ";
          }
          cnt += n / m;
        }
        cout << "\n";
      }
      for (int j = 1; j <= (n % m) * (n / m + 1); j++) {
        long long qwe = a[0];
        a.pop_front();
        a.push_back(qwe);
      }
    }
  }
}

### Prompt

In cpp, your task is to solve 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>
const long long N = 1e6 + 10;
const long long INF = 1e18;
const long long inf = -1e18;
const long long MOD = 998244353;
using namespace std;
long long n, m, k;
long long inv(long long a, long long m) {
  long long temp = m, q, t, u = 0, v = 1;
  if (m == 1) return 0;
  while (a > 1) {
    q = a / m;
    t = m;
    m = a % m;
    a = t;
    t = u;
    u = v - q * u;
    v = t;
  }
  if (v < 0) v += temp;
  return v;
}
int main() {
  ios_base::sync_with_stdio(0);
  int o = 1;
  cin >> o;
  while (o--) {
    cin >> n >> m >> k;
    deque<long long> a;
    for (int i = 0; i < n; i++) {
      a.push_back(i + 1);
    }
    for (int i = 1; i <= k; i++) {
      long long cnt = 1;
      for (int j = 1; j <= m; j++) {
        if (j <= n % m) {
          cout << n / m + 1 << " ";
          for (int j = cnt; j <= cnt + n / m + 1 - 1; j++) {
            cout << a[j - 1] << " ";
          }
          cnt += n / m + 1;
        } else {
          cout << n / m << " ";
          for (int j = cnt; j <= cnt + n / m - 1; j++) {
            cout << a[j - 1] << " ";
          }
          cnt += n / m;
        }
        cout << "\n";
      }
      for (int j = 1; j <= (n % m) * (n / m + 1); j++) {
        long long qwe = a[0];
        a.pop_front();
        a.push_back(qwe);
      }
    }
  }
}

```