#include <bits/stdc++.h>
using namespace std;
const long long int ll_MAX = 1e14 + 5;
const long long int MOD = 998244353;
long long int fast_expo(long long int x, long long int p) {
  if (p == 0)
    return 1;
  else if (p % 2 == 0) {
    long long int t = fast_expo(x, p / 2) % MOD;
    return (t * t) % MOD;
  } else
    return (x * (fast_expo(x, p - 1)) % MOD) % MOD;
}
long long int gcd(long long int a, long long int b) {
  if (a == 0 || b == 0)
    return a + b;
  else
    return gcd(b % a, a);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long int t_c;
  t_c = 1;
  cin >> t_c;
  for (long long int t_i = 0; t_i < t_c; ++t_i) {
    long long int n, m, k;
    cin >> n >> m >> k;
    long long int curr = 1;
    long long int val1 = n / m, cnt1 = m - n % m;
    long long int val2 = n / m + 1, cnt2 = n % m;
    for (long long int i = 1; i <= k; ++i) {
      long long int tmp = curr;
      for (long long int j = 1; j <= m; ++j) {
        if (j <= cnt2) {
          cout << val2 << " ";
          for (long long int x = 1; x <= val2; ++x) {
            cout << tmp << " ";
            tmp = tmp % n + 1;
          }
          curr = tmp;
        } else {
          cout << val1 << " ";
          for (long long int x = 1; x <= val1; ++x) {
            cout << tmp << " ";
            tmp = tmp % n + 1;
          }
        }
        cout << "\n";
      }
    }
    cout << "\n";
  }
  return 0;
}

### 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;
const long long int ll_MAX = 1e14 + 5;
const long long int MOD = 998244353;
long long int fast_expo(long long int x, long long int p) {
  if (p == 0)
    return 1;
  else if (p % 2 == 0) {
    long long int t = fast_expo(x, p / 2) % MOD;
    return (t * t) % MOD;
  } else
    return (x * (fast_expo(x, p - 1)) % MOD) % MOD;
}
long long int gcd(long long int a, long long int b) {
  if (a == 0 || b == 0)
    return a + b;
  else
    return gcd(b % a, a);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long int t_c;
  t_c = 1;
  cin >> t_c;
  for (long long int t_i = 0; t_i < t_c; ++t_i) {
    long long int n, m, k;
    cin >> n >> m >> k;
    long long int curr = 1;
    long long int val1 = n / m, cnt1 = m - n % m;
    long long int val2 = n / m + 1, cnt2 = n % m;
    for (long long int i = 1; i <= k; ++i) {
      long long int tmp = curr;
      for (long long int j = 1; j <= m; ++j) {
        if (j <= cnt2) {
          cout << val2 << " ";
          for (long long int x = 1; x <= val2; ++x) {
            cout << tmp << " ";
            tmp = tmp % n + 1;
          }
          curr = tmp;
        } else {
          cout << val1 << " ";
          for (long long int x = 1; x <= val1; ++x) {
            cout << tmp << " ";
            tmp = tmp % n + 1;
          }
        }
        cout << "\n";
      }
    }
    cout << "\n";
  }
  return 0;
}

```