#include <bits/stdc++.h>
using namespace std;
long long binpow(long long base, long long ex,
                 long long mod = (long long)1000000007) {
  long long ans = 1;
  base = base % mod;
  if (base == 0) {
    return 0;
  }
  while (ex > 0) {
    if (ex % 2 == 1) {
      ans = (ans * base) % mod;
    }
    base = (base * base) % mod;
    ex = ex / 2;
  }
  return ans;
}
void read(vector<long long> &arr, long long n) {
  if (arr.size() != n) {
    arr.assign(n, 0);
  }
  for (int i = 0; i < n; i++) cin >> arr[i];
}
inline long long min(long long a, long long b) {
  if (a > b) return b;
  return a;
}
inline long long max(long long a, long long b) {
  if (a > b) return a;
  return b;
}
inline long long dif(long long a, long long b) {
  if (a > b) return a - b;
  return b - a;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long lcm(long long a, long long b) {
  long long k = gcd(a, b);
  return (a * b) / k;
}
void solve() {
  long long n, m, k;
  cin >> n >> m >> k;
  vector<vector<vector<long long>>> ans(k, vector<vector<long long>>(2));
  long long quo = n / m;
  long long rem = n % m;
  long long constant = k * ((n % m) * (n / m + 1));
  long long x = ((constant) + (n - (constant) % n)) / n - 1;
  long long a = (constant % n);
  vector<long long> arr(n, x);
  for (long long i = 0; (i) < long(a); (i)++) {
    arr[i]++;
  }
  long long last = 0;
  for (int curr = 0; curr <= 4 * n; curr++) {
    int i = curr % n;
    long long new_last = last;
    for (int game = 0; game < k; game++) {
      if (arr[i] and ans[(game + last) % k][0].size() < (quo + 1) * rem) {
        arr[i]--;
        new_last = (game + last) % k;
        ans[(game + last) % k][0].push_back(i + 1);
      } else if (ans[(game + last) % k][1].size() < (quo) * (n - rem)) {
        ans[(game + last) % k][1].push_back(i + 1);
      }
    }
    last = (new_last + 1) % k;
  }
  for (int game = 0; game < k; game++) {
    int up = 0;
    for (int count = 0; count < rem; count++) {
      cout << quo + 1 << " ";
      for (int i = 0; i <= quo; i++) {
        cout << ans[game][0][up] << " ";
        up++;
      }
      cout << endl;
    }
    int down = 0;
    for (int count = 0; count < (m - rem); count++) {
      cout << quo << " ";
      for (int i = 0; i < quo; i++) {
        cout << ans[game][1][down] << " ";
        down++;
      }
      cout << endl;
    }
  }
  cout << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout << fixed << showpoint;
  cout << setprecision(12);
  long long test_m = 1;
  cin >> test_m;
  while (test_m--) {
    solve();
  }
}

### Prompt

Your task is to create 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;
long long binpow(long long base, long long ex,
                 long long mod = (long long)1000000007) {
  long long ans = 1;
  base = base % mod;
  if (base == 0) {
    return 0;
  }
  while (ex > 0) {
    if (ex % 2 == 1) {
      ans = (ans * base) % mod;
    }
    base = (base * base) % mod;
    ex = ex / 2;
  }
  return ans;
}
void read(vector<long long> &arr, long long n) {
  if (arr.size() != n) {
    arr.assign(n, 0);
  }
  for (int i = 0; i < n; i++) cin >> arr[i];
}
inline long long min(long long a, long long b) {
  if (a > b) return b;
  return a;
}
inline long long max(long long a, long long b) {
  if (a > b) return a;
  return b;
}
inline long long dif(long long a, long long b) {
  if (a > b) return a - b;
  return b - a;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long lcm(long long a, long long b) {
  long long k = gcd(a, b);
  return (a * b) / k;
}
void solve() {
  long long n, m, k;
  cin >> n >> m >> k;
  vector<vector<vector<long long>>> ans(k, vector<vector<long long>>(2));
  long long quo = n / m;
  long long rem = n % m;
  long long constant = k * ((n % m) * (n / m + 1));
  long long x = ((constant) + (n - (constant) % n)) / n - 1;
  long long a = (constant % n);
  vector<long long> arr(n, x);
  for (long long i = 0; (i) < long(a); (i)++) {
    arr[i]++;
  }
  long long last = 0;
  for (int curr = 0; curr <= 4 * n; curr++) {
    int i = curr % n;
    long long new_last = last;
    for (int game = 0; game < k; game++) {
      if (arr[i] and ans[(game + last) % k][0].size() < (quo + 1) * rem) {
        arr[i]--;
        new_last = (game + last) % k;
        ans[(game + last) % k][0].push_back(i + 1);
      } else if (ans[(game + last) % k][1].size() < (quo) * (n - rem)) {
        ans[(game + last) % k][1].push_back(i + 1);
      }
    }
    last = (new_last + 1) % k;
  }
  for (int game = 0; game < k; game++) {
    int up = 0;
    for (int count = 0; count < rem; count++) {
      cout << quo + 1 << " ";
      for (int i = 0; i <= quo; i++) {
        cout << ans[game][0][up] << " ";
        up++;
      }
      cout << endl;
    }
    int down = 0;
    for (int count = 0; count < (m - rem); count++) {
      cout << quo << " ";
      for (int i = 0; i < quo; i++) {
        cout << ans[game][1][down] << " ";
        down++;
      }
      cout << endl;
    }
  }
  cout << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout << fixed << showpoint;
  cout << setprecision(12);
  long long test_m = 1;
  cin >> test_m;
  while (test_m--) {
    solve();
  }
}

```