#include <bits/stdc++.h>
using namespace std;
vector<long long> prime;
vector<bool> mark(1e6 + 6, true);
vector<long long> fact(1e6 + 1, 1);
vector<long long> inv(1e6 + 1, 1);
vector<long long> invf(1e6 + 1, 1);
long long Mod = 1e9 + 7;
void seive(long long n) {
  long long i, j;
  mark[1] = false;
  prime.push_back(2);
  for (i = 4; i <= n; i += 2) mark[i] = false;
  for (i = 3; i * i <= n; i++) {
    if (mark[i]) {
      for (j = i * i; j <= n; j += 2 * i) {
        mark[j] = false;
      }
    }
  }
  for (i = 3; i <= n; i += 2)
    if (mark[i]) prime.push_back(i);
}
long long ck(long long n) {
  long long l = 1, r = n, mid;
  while (l <= r) {
    mid = (l + r) / 2;
    if ((mid * (mid + 1)) / 2 < n)
      l = mid + 1;
    else
      r = mid - 1;
  }
  return l;
}
void ft() {
  for (long long i = 2; i <= 1e6; i++) {
    fact[i] = (i * fact[i - 1]) % Mod;
  }
}
void modinv() {
  inv[0] = inv[1] = 1;
  for (long long i = 2; i <= 1e6; i++) {
    inv[i] = inv[Mod % i] * (Mod - Mod / i) % Mod;
  }
}
void modfactinv() {
  invf[0] = invf[1] = 1;
  for (long long i = 2; i <= 1e6; i++) {
    invf[i] = (invf[i - 1] * inv[i]) % Mod;
  }
}
void solve(long long cs) {
  long long n, m, i, z, h, w, e, k = 0, j, a, x, y, b, ans = 1;
  string s;
  cin >> n >> m >> k;
  x = n / m;
  y = x;
  if (n % m) x++;
  w = n % m;
  for (z = 0; z < k; z++) {
    for (i = 0; i < m; i++) {
      if (i < w) {
        cout << x << " ";
        for (j = 0; j < x; j++) cout << (j + (z * w + i) * x) % n + 1 << " ";
        cout << '\n';
      } else {
        cout << y << " ";
        for (j = 0; j < y; j++)
          cout << (j + (i - w) * y + ((z + 1) * w) * x) % n + 1 << " ";
        cout << '\n';
      }
    }
  }
  cout << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t = 1, cs = 1;
  cin >> t;
  while (t--) {
    solve(cs++);
  }
}

### Prompt

Generate 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;
vector<long long> prime;
vector<bool> mark(1e6 + 6, true);
vector<long long> fact(1e6 + 1, 1);
vector<long long> inv(1e6 + 1, 1);
vector<long long> invf(1e6 + 1, 1);
long long Mod = 1e9 + 7;
void seive(long long n) {
  long long i, j;
  mark[1] = false;
  prime.push_back(2);
  for (i = 4; i <= n; i += 2) mark[i] = false;
  for (i = 3; i * i <= n; i++) {
    if (mark[i]) {
      for (j = i * i; j <= n; j += 2 * i) {
        mark[j] = false;
      }
    }
  }
  for (i = 3; i <= n; i += 2)
    if (mark[i]) prime.push_back(i);
}
long long ck(long long n) {
  long long l = 1, r = n, mid;
  while (l <= r) {
    mid = (l + r) / 2;
    if ((mid * (mid + 1)) / 2 < n)
      l = mid + 1;
    else
      r = mid - 1;
  }
  return l;
}
void ft() {
  for (long long i = 2; i <= 1e6; i++) {
    fact[i] = (i * fact[i - 1]) % Mod;
  }
}
void modinv() {
  inv[0] = inv[1] = 1;
  for (long long i = 2; i <= 1e6; i++) {
    inv[i] = inv[Mod % i] * (Mod - Mod / i) % Mod;
  }
}
void modfactinv() {
  invf[0] = invf[1] = 1;
  for (long long i = 2; i <= 1e6; i++) {
    invf[i] = (invf[i - 1] * inv[i]) % Mod;
  }
}
void solve(long long cs) {
  long long n, m, i, z, h, w, e, k = 0, j, a, x, y, b, ans = 1;
  string s;
  cin >> n >> m >> k;
  x = n / m;
  y = x;
  if (n % m) x++;
  w = n % m;
  for (z = 0; z < k; z++) {
    for (i = 0; i < m; i++) {
      if (i < w) {
        cout << x << " ";
        for (j = 0; j < x; j++) cout << (j + (z * w + i) * x) % n + 1 << " ";
        cout << '\n';
      } else {
        cout << y << " ";
        for (j = 0; j < y; j++)
          cout << (j + (i - w) * y + ((z + 1) * w) * x) % n + 1 << " ";
        cout << '\n';
      }
    }
  }
  cout << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t = 1, cs = 1;
  cin >> t;
  while (t--) {
    solve(cs++);
  }
}

```