#include <bits/stdc++.h>
using namespace std;
using ll = long long;
template <class T, class U>
inline bool chmax(T &a, U b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <class T, class U>
inline bool chmin(T &a, U b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
constexpr int INF = 1000000000;
constexpr ll llINF = 3000000000000000000;
constexpr int mod = 998244353;
constexpr double eps = 1e-10;
vector<int> prime_list(int n) {
  vector<int> v(n + 1), res;
  for (int i = n; i >= 2; i--) {
    for (int j = i; j <= n; j += i) v[j] = i;
  }
  for (int i = 2; i <= n; i++) {
    if (v[i] == i) res.push_back(i);
  }
  return res;
}
vector<int> next_divisor(int n) {
  vector<int> v(n + 1);
  for (int i = n; i >= 2; i--) {
    for (int j = i; j <= n; j += i) v[j] = i;
  }
  return v;
}
ll modpow(ll a, ll b, ll m = mod) {
  ll res = 1;
  while (b) {
    if (b & 1) {
      res *= a;
      res %= m;
    }
    a *= a;
    a %= m;
    b >>= 1;
  }
  return res;
}
vector<ll> inv, fact, factinv;
void init_fact(int n) {
  inv.resize(n + 1);
  fact.resize(n + 1);
  factinv.resize(n + 1);
  inv[0] = 0;
  inv[1] = 1;
  fact[0] = 1;
  factinv[0] = 1;
  for (ll i = 1; i <= n; i++) {
    if (i >= 2) inv[i] = mod - ((mod / i) * inv[mod % i] % mod);
    fact[i] = (fact[i - 1] * i) % mod;
    factinv[i] = factinv[i - 1] * inv[i] % mod;
  }
}
ll modinv(ll a, ll m = mod) {
  ll b = m, u = 1, v = 0;
  while (b) {
    ll t = a / b;
    a -= t * b;
    swap(a, b);
    u -= t * v;
    swap(u, v);
  }
  u %= m;
  if (u < 0) u += m;
  return u;
}
ll comb(int a, int b) {
  if (a < b) return 0;
  if (a < 0) return 0;
  return fact[a] * factinv[a - b] % mod * factinv[b] % mod;
}
void solve() {
  int n, m, k;
  cin >> n >> m >> k;
  int st = 0, ac = 0;
  for (int i = 0; i < k; i++) {
    st = ac;
    for (int j = 0; j < m; j++) {
      if (j < n % m) {
        cout << n / m + 1;
        for (int _ = 0; _ < n / m + 1; _++) {
          cout << " " << st + 1;
          st++;
          if (st == n) st = 0;
        }
        ac = st;
      } else {
        cout << n / m;
        for (int _ = 0; _ < n / m; _++) {
          cout << " " << st + 1;
          st++;
          if (st == n) st = 0;
        }
      }
      cout << endl;
    }
  }
  cout << endl;
}
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int t;
  cin >> t;
  while (t--) solve();
}

### Prompt

Your challenge is to write 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;
using ll = long long;
template <class T, class U>
inline bool chmax(T &a, U b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <class T, class U>
inline bool chmin(T &a, U b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
constexpr int INF = 1000000000;
constexpr ll llINF = 3000000000000000000;
constexpr int mod = 998244353;
constexpr double eps = 1e-10;
vector<int> prime_list(int n) {
  vector<int> v(n + 1), res;
  for (int i = n; i >= 2; i--) {
    for (int j = i; j <= n; j += i) v[j] = i;
  }
  for (int i = 2; i <= n; i++) {
    if (v[i] == i) res.push_back(i);
  }
  return res;
}
vector<int> next_divisor(int n) {
  vector<int> v(n + 1);
  for (int i = n; i >= 2; i--) {
    for (int j = i; j <= n; j += i) v[j] = i;
  }
  return v;
}
ll modpow(ll a, ll b, ll m = mod) {
  ll res = 1;
  while (b) {
    if (b & 1) {
      res *= a;
      res %= m;
    }
    a *= a;
    a %= m;
    b >>= 1;
  }
  return res;
}
vector<ll> inv, fact, factinv;
void init_fact(int n) {
  inv.resize(n + 1);
  fact.resize(n + 1);
  factinv.resize(n + 1);
  inv[0] = 0;
  inv[1] = 1;
  fact[0] = 1;
  factinv[0] = 1;
  for (ll i = 1; i <= n; i++) {
    if (i >= 2) inv[i] = mod - ((mod / i) * inv[mod % i] % mod);
    fact[i] = (fact[i - 1] * i) % mod;
    factinv[i] = factinv[i - 1] * inv[i] % mod;
  }
}
ll modinv(ll a, ll m = mod) {
  ll b = m, u = 1, v = 0;
  while (b) {
    ll t = a / b;
    a -= t * b;
    swap(a, b);
    u -= t * v;
    swap(u, v);
  }
  u %= m;
  if (u < 0) u += m;
  return u;
}
ll comb(int a, int b) {
  if (a < b) return 0;
  if (a < 0) return 0;
  return fact[a] * factinv[a - b] % mod * factinv[b] % mod;
}
void solve() {
  int n, m, k;
  cin >> n >> m >> k;
  int st = 0, ac = 0;
  for (int i = 0; i < k; i++) {
    st = ac;
    for (int j = 0; j < m; j++) {
      if (j < n % m) {
        cout << n / m + 1;
        for (int _ = 0; _ < n / m + 1; _++) {
          cout << " " << st + 1;
          st++;
          if (st == n) st = 0;
        }
        ac = st;
      } else {
        cout << n / m;
        for (int _ = 0; _ < n / m; _++) {
          cout << " " << st + 1;
          st++;
          if (st == n) st = 0;
        }
      }
      cout << endl;
    }
  }
  cout << endl;
}
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int t;
  cin >> t;
  while (t--) solve();
}

```