#include <bits/stdc++.h>
using namespace std;
template <typename T>
bool inline chkMin(T &x, T y) {
  return (y < x) ? x = y, 1 : 0;
}
template <typename T>
bool inline chkMax(T &x, T y) {
  return (y > x) ? x = y, 1 : 0;
}
template <typename T>
void inline read(T &x) {
  x = 0;
  int f = 1;
  char s = getchar();
  while (s > '9' || s < '0') {
    if (s == '-') f = -1;
    s = getchar();
  }
  while (s <= '9' && s >= '0') x = x * 10 + s - '0', s = getchar();
  x *= f;
}
const int N = 2e5 + 5;
int n, m, k, p[N], w[N];
vector<int> b[N];
int main() {
  int T;
  read(T);
  while (T--) {
    read(n), read(m), read(k);
    int t = n % m;
    for (int i = 0; i < n; i++) p[i] = i;
    int A = n / m, B = (n + m - 1) / m;
    for (int i = 0; i < k; i++) {
      int id = 0, now = 0;
      for (int i = 0; i < t; i++) {
        for (int j = 0; j < B; j++) {
          b[now].push_back(p[id++]);
        }
        now++;
      }
      for (int i = 0; i < m - t; i++) {
        for (int j = 0; j < A; j++) {
          b[now].push_back(p[id++]);
        }
        now++;
      }
      for (int j = 0; j < m; j++) {
        printf("%lu", b[j].size());
        for (int v : b[j]) {
          printf(" %d", v + 1);
        }
        puts("");
      }
      for (int j = 0; j < m; j++) b[j].clear();
      for (int u = 0; u < n; u++) w[u] = p[u];
      for (int u = 0; u < n; u++) p[u] = w[(u + B * t) % n];
    }
    puts("");
  }
  return 0;
}

### Prompt

Please formulate 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;
template <typename T>
bool inline chkMin(T &x, T y) {
  return (y < x) ? x = y, 1 : 0;
}
template <typename T>
bool inline chkMax(T &x, T y) {
  return (y > x) ? x = y, 1 : 0;
}
template <typename T>
void inline read(T &x) {
  x = 0;
  int f = 1;
  char s = getchar();
  while (s > '9' || s < '0') {
    if (s == '-') f = -1;
    s = getchar();
  }
  while (s <= '9' && s >= '0') x = x * 10 + s - '0', s = getchar();
  x *= f;
}
const int N = 2e5 + 5;
int n, m, k, p[N], w[N];
vector<int> b[N];
int main() {
  int T;
  read(T);
  while (T--) {
    read(n), read(m), read(k);
    int t = n % m;
    for (int i = 0; i < n; i++) p[i] = i;
    int A = n / m, B = (n + m - 1) / m;
    for (int i = 0; i < k; i++) {
      int id = 0, now = 0;
      for (int i = 0; i < t; i++) {
        for (int j = 0; j < B; j++) {
          b[now].push_back(p[id++]);
        }
        now++;
      }
      for (int i = 0; i < m - t; i++) {
        for (int j = 0; j < A; j++) {
          b[now].push_back(p[id++]);
        }
        now++;
      }
      for (int j = 0; j < m; j++) {
        printf("%lu", b[j].size());
        for (int v : b[j]) {
          printf(" %d", v + 1);
        }
        puts("");
      }
      for (int j = 0; j < m; j++) b[j].clear();
      for (int u = 0; u < n; u++) w[u] = p[u];
      for (int u = 0; u < n; u++) p[u] = w[(u + B * t) % n];
    }
    puts("");
  }
  return 0;
}

```