#include <bits/stdc++.h>
using namespace std;
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    int n, m, k;
    scanf("%d%d%d", &n, &m, &k);
    int rem = (n % m) * ((n + m - 1) / m);
    int peo = (n + m - 1) / m;
    if (n % m == 0) {
      rem = n;
      peo = n / m;
    }
    int now = 0;
    while (k--) {
      int sum = 0;
      int cnt = 0;
      for (int i = now; i < now + rem; i++) {
        if (sum % peo == 0) printf("%d ", peo);
        printf("%d ", i % n + 1);
        cnt++;
        sum++;
        if (cnt % peo == 0) {
          printf("\n");
        }
      }
      sum = 0;
      cnt = 0;
      for (int i = now + rem; i <= now + n - 1; i++) {
        if (sum % (peo - 1) == 0) printf("%d ", peo - 1);
        printf("%d ", i % n + 1);
        cnt++;
        sum++;
        if (cnt % (peo - 1) == 0) printf("\n");
      }
      now = (now + rem) % n;
    }
  }
}

### Prompt

Your challenge is to write 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;
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    int n, m, k;
    scanf("%d%d%d", &n, &m, &k);
    int rem = (n % m) * ((n + m - 1) / m);
    int peo = (n + m - 1) / m;
    if (n % m == 0) {
      rem = n;
      peo = n / m;
    }
    int now = 0;
    while (k--) {
      int sum = 0;
      int cnt = 0;
      for (int i = now; i < now + rem; i++) {
        if (sum % peo == 0) printf("%d ", peo);
        printf("%d ", i % n + 1);
        cnt++;
        sum++;
        if (cnt % peo == 0) {
          printf("\n");
        }
      }
      sum = 0;
      cnt = 0;
      for (int i = now + rem; i <= now + n - 1; i++) {
        if (sum % (peo - 1) == 0) printf("%d ", peo - 1);
        printf("%d ", i % n + 1);
        cnt++;
        sum++;
        if (cnt % (peo - 1) == 0) printf("\n");
      }
      now = (now + rem) % n;
    }
  }
}

```