#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
using ull = unsigned long long;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
using pld = pair<ld, ld>;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int d4x[4] = {1, 0, -1, 0};
int d4y[4] = {0, 1, 0, -1};
int d8x[8] = {0, 1, 1, 1, 0, -1, -1, -1};
int d8y[8] = {1, 1, 0, -1, -1, -1, 0, 1};
template <class X, class Y>
bool minimize(X &x, const Y &y) {
  if (x > y) {
    x = y;
    return true;
  }
  return false;
}
template <class X, class Y>
bool maximize(X &x, const Y &y) {
  if (x < y) {
    x = y;
    return true;
  }
  return false;
}
const int MOD = 1e9 + 7;
template <class X, class Y>
void add(X &x, const Y &y) {
  x = (x + y);
  if (x >= MOD) x -= MOD;
}
template <class X, class Y>
void sub(X &x, const Y &y) {
  x = (x - y);
  if (x < 0) x += MOD;
}
const ll INF = 1e9;
const int N = 5e5 + 10;
int cnt[N];
void solve() {
  int n, m, k;
  cin >> n >> m >> k;
  int each = n / m;
  int r = n % m;
  struct T {
    int index, c;
    bool operator<(const T &temp) const { return c > temp.c; };
  };
  priority_queue<T> q;
  for (int i = 1; i <= n; i++) {
    cnt[i] = 0;
    q.push({i, 0});
  }
  for (int i = 1; i <= k; i++) {
    for (int j = 1; j <= r; j++) {
      cout << each + 1 << " ";
      for (int t = 1; t <= each + 1; t++) {
        T curr = q.top();
        q.pop();
        cnt[curr.index]++;
        cout << curr.index << " ";
      }
      cout << '\n';
    }
    for (int j = r + 1; j <= m; j++) {
      cout << each << " ";
      for (int t = 1; t <= each; t++) {
        T curr = q.top();
        q.pop();
        cout << curr.index << " ";
      }
      cout << '\n';
    }
    for (int j = 1; j <= n; j++) q.push({j, cnt[j]});
  }
  cout << '\n';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int tc = 1, ddd = 0;
  cin >> tc;
  while (tc--) {
    solve();
  }
}

### 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>
using namespace std;
using ll = long long;
using ld = long double;
using ull = unsigned long long;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
using pld = pair<ld, ld>;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int d4x[4] = {1, 0, -1, 0};
int d4y[4] = {0, 1, 0, -1};
int d8x[8] = {0, 1, 1, 1, 0, -1, -1, -1};
int d8y[8] = {1, 1, 0, -1, -1, -1, 0, 1};
template <class X, class Y>
bool minimize(X &x, const Y &y) {
  if (x > y) {
    x = y;
    return true;
  }
  return false;
}
template <class X, class Y>
bool maximize(X &x, const Y &y) {
  if (x < y) {
    x = y;
    return true;
  }
  return false;
}
const int MOD = 1e9 + 7;
template <class X, class Y>
void add(X &x, const Y &y) {
  x = (x + y);
  if (x >= MOD) x -= MOD;
}
template <class X, class Y>
void sub(X &x, const Y &y) {
  x = (x - y);
  if (x < 0) x += MOD;
}
const ll INF = 1e9;
const int N = 5e5 + 10;
int cnt[N];
void solve() {
  int n, m, k;
  cin >> n >> m >> k;
  int each = n / m;
  int r = n % m;
  struct T {
    int index, c;
    bool operator<(const T &temp) const { return c > temp.c; };
  };
  priority_queue<T> q;
  for (int i = 1; i <= n; i++) {
    cnt[i] = 0;
    q.push({i, 0});
  }
  for (int i = 1; i <= k; i++) {
    for (int j = 1; j <= r; j++) {
      cout << each + 1 << " ";
      for (int t = 1; t <= each + 1; t++) {
        T curr = q.top();
        q.pop();
        cnt[curr.index]++;
        cout << curr.index << " ";
      }
      cout << '\n';
    }
    for (int j = r + 1; j <= m; j++) {
      cout << each << " ";
      for (int t = 1; t <= each; t++) {
        T curr = q.top();
        q.pop();
        cout << curr.index << " ";
      }
      cout << '\n';
    }
    for (int j = 1; j <= n; j++) q.push({j, cnt[j]});
  }
  cout << '\n';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int tc = 1, ddd = 0;
  cin >> tc;
  while (tc--) {
    solve();
  }
}

```