#include <bits/stdc++.h>
#pragma warning(disable : 4996)
using namespace std;
const long long MOD = 1e9 + 7;
const long long MODD = 1e9 + 1;
const long long MODL = 998244353;
const long long N = 200005;
const long long INF = 1e18;
const long long mx = 1e5 + 1;
long long Power(long long x, long long y) {
  long long r = 1;
  while (y) {
    if (y & 1) r = 1ll * r * x % MOD;
    x = 1ll * x * x % MOD, y >>= 1;
  }
  return r;
}
bool sortbysec(const pair<int, int>& a, const pair<int, int>& b) {
  return (a.second < b.second);
}
bool sortstring(const string& a, const string& b) { return a + b < b + a; }
long long GCD(long long a, long long b) {
  if (b == 0) return a;
  a %= b;
  return GCD(b, a);
}
bool isZero(int i) { return i == 0; }
int ce(int n, int m) {
  int ans = 0;
  if (n % m == 0)
    ans = n / m;
  else
    ans = n / m + 1;
  return ans;
}
int solve(int n) {
  int m, k;
  cin >> m >> k;
  int p = 0;
  int q = 0, qq = 0;
  int r = 0;
  int x = ce(n, m);
  int s = 0;
  if (n % m != 0) {
    for (int i = n - 1; i >= 0; i--) {
      if (i % x == 0 && (n - i) % (n / m) == 0 &&
          (i / x + (n - i) / (n / m) == m)) {
        s = i;
        break;
      }
    }
  }
  s = (n - s) / (n / m);
  for (int i = 0; i < k; i++) {
    for (int j = 0; j < m - s; j++) {
      cout << x << " ";
      r = j * x + p;
      for (int k = 0; k < x; k++) {
        cout << r % n + 1 << " ";
        q = r % n;
        r++;
      }
      cout << "\n";
    }
    qq = q;
    for (int j = 0; j < s; j++) {
      cout << n / m << " ";
      r = q + 1;
      for (int k = 0; k < n / m; k++) {
        cout << r % n + 1 << " ";
        q = r % n;
        r++;
      }
      cout << "\n";
    }
    p = qq + 1;
  }
  cout << "\n";
  return 0;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  int n = 0;
  cin >> t;
  while (t--) {
    cin >> n;
    solve(n);
  }
  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>
#pragma warning(disable : 4996)
using namespace std;
const long long MOD = 1e9 + 7;
const long long MODD = 1e9 + 1;
const long long MODL = 998244353;
const long long N = 200005;
const long long INF = 1e18;
const long long mx = 1e5 + 1;
long long Power(long long x, long long y) {
  long long r = 1;
  while (y) {
    if (y & 1) r = 1ll * r * x % MOD;
    x = 1ll * x * x % MOD, y >>= 1;
  }
  return r;
}
bool sortbysec(const pair<int, int>& a, const pair<int, int>& b) {
  return (a.second < b.second);
}
bool sortstring(const string& a, const string& b) { return a + b < b + a; }
long long GCD(long long a, long long b) {
  if (b == 0) return a;
  a %= b;
  return GCD(b, a);
}
bool isZero(int i) { return i == 0; }
int ce(int n, int m) {
  int ans = 0;
  if (n % m == 0)
    ans = n / m;
  else
    ans = n / m + 1;
  return ans;
}
int solve(int n) {
  int m, k;
  cin >> m >> k;
  int p = 0;
  int q = 0, qq = 0;
  int r = 0;
  int x = ce(n, m);
  int s = 0;
  if (n % m != 0) {
    for (int i = n - 1; i >= 0; i--) {
      if (i % x == 0 && (n - i) % (n / m) == 0 &&
          (i / x + (n - i) / (n / m) == m)) {
        s = i;
        break;
      }
    }
  }
  s = (n - s) / (n / m);
  for (int i = 0; i < k; i++) {
    for (int j = 0; j < m - s; j++) {
      cout << x << " ";
      r = j * x + p;
      for (int k = 0; k < x; k++) {
        cout << r % n + 1 << " ";
        q = r % n;
        r++;
      }
      cout << "\n";
    }
    qq = q;
    for (int j = 0; j < s; j++) {
      cout << n / m << " ";
      r = q + 1;
      for (int k = 0; k < n / m; k++) {
        cout << r % n + 1 << " ";
        q = r % n;
        r++;
      }
      cout << "\n";
    }
    p = qq + 1;
  }
  cout << "\n";
  return 0;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  int n = 0;
  cin >> t;
  while (t--) {
    cin >> n;
    solve(n);
  }
  return 0;
}

```