#include <bits/stdc++.h>
using namespace std;
const long long MOD = pow(10, 9) + 7;
long long fac[1001];
template <typename T, typename U>
static inline void amin(T &x, U y) {
  if (y < x) x = y;
}
template <typename T, typename U>
static inline void amax(T &x, U y) {
  if (x < y) x = y;
}
void SieveOfEratosthenes(long long n) {
  bool prime[n + 1];
  memset(prime, true, sizeof(prime));
  for (long long p = 2; p * p <= n; p++) {
    if (prime[p] == true) {
      for (long long i = p * p; i <= n; i += p) prime[i] = false;
    }
  }
  for (long long p = 2; p <= n; p++)
    if (prime[p]) {
    }
}
long long power(long long x, long long y, long long p) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long modInverse(long long n, long long p) { return power(n, p - 2, p); }
long long nCrModPFermat(long long n, long long r, long long p) {
  if (n < r) return 0;
  if (r == 0) return 1;
  return (fac[n] * modInverse(fac[r], p) % p * modInverse(fac[n - r], p) % p) %
         p;
}
void dfs(long long v, vector<vector<long long>> &adj,
         vector<long long> &visited) {
  for (long long u : adj[v]) {
    if (!visited[u]) {
      visited[u] = 1;
    }
  }
}
long long parent[300002], sz[300002];
long long find_set(long long v) {
  if (v == parent[v]) return v;
  return parent[v] = find_set(parent[v]);
}
void make_set(long long v) {
  parent[v] = v;
  sz[v] = 1;
}
void union_sets(long long a, long long b) {
  a = find_set(a);
  b = find_set(b);
  if (a != b) {
    if (sz[a] < sz[b]) swap(a, b);
    parent[b] = a;
    sz[a] += sz[b];
  }
}
class segTree {
 public:
  long long len;
  vector<long long> a;
  vector<pair<long long, long long>> t;
  segTree(vector<long long> &val) {
    len = val.size();
    a.resize(len);
    long long i;
    a = val;
    if ((len & (len - 1)) != 0) {
      long long np = ceil(log2(len));
      long long pval = pow(2, np);
      long long diff = pval - len;
      len = pval;
      for (i = 0; i < diff; i++) a.push_back(pow(10, 10));
    }
    t.resize(4 * len + 1);
  }
  void build(long long idx, long long tl, long long tr) {
    if (tl == tr) {
      t[idx] = {a[tl], 1};
      return;
    }
    long long tm = (tl + tr) / 2;
    build(idx * 2, tl, tm);
    build(idx * 2 + 1, tm + 1, tr);
    long long count = 0;
    long long mn = min(t[2 * idx].first, t[2 * idx + 1].first);
    if (mn == t[2 * idx].first) count += t[2 * idx].second;
    if (mn == t[2 * idx + 1].first) count += t[2 * idx + 1].second;
    t[idx] = {mn, count};
  }
  pair<long long, long long> query(long long idx, long long tl, long long tr,
                                   long long l, long long r) {
    if (l > r) return {pow(10, 10), 1};
    if (l == tl && r == tr) {
      return t[idx];
    }
    long long tm = (tl + tr) / 2;
    pair<long long, long long> res1 = query(idx * 2, tl, tm, l, min(r, tm));
    pair<long long, long long> res2 =
        query(idx * 2 + 1, tm + 1, tr, max(l, tm + 1), r);
    if (res1.first < res2.first) {
      return res1;
    } else if (res2.first < res1.first) {
      return res2;
    } else {
      return {res1.first, res1.second + res2.second};
    }
  }
  void update(long long idx, long long tl, long long tr, long long pos,
              long long val) {
    if (tl == tr) {
      t[idx] = {val, 1};
      return;
    }
    long long tm = (tl + tr) / 2;
    if (pos <= tm)
      update(2 * idx, tl, tm, pos, val);
    else
      update(2 * idx + 1, tm + 1, tr, pos, val);
    long long count = 0;
    long long mn = min(t[2 * idx].first, t[2 * idx + 1].first);
    if (mn == t[2 * idx].first) count += t[2 * idx].second;
    if (mn == t[2 * idx + 1].first) count += t[2 * idx + 1].second;
    t[idx] = {mn, count};
  }
};
void solve() {
  long long i, j, n, m, k;
  cin >> n >> m >> k;
  if (n % m == 0) {
    for (i = 1; i <= k; i++) {
      for (j = 1; j <= m; j++) {
        long long z = n / m;
        cout << z << " ";
        long long s = z * (j - 1) + 1;
        long long e = z * j;
        for (int r = s; r <= e; r++) {
          cout << r << " ";
        }
        cout << '\n';
      }
    }
    return;
  }
  long long big = n % m, sm = m - big;
  long long c = 1;
  for (i = 1; i <= k; i++) {
    long long z = n / m;
    for (j = 1; j <= big; j++) {
      cout << z + 1 << " ";
      for (long long r = 1; r <= z + 1; r++) {
        cout << c << " ";
        c++;
        if (c > n) c = 1;
      }
      cout << '\n';
    }
    long long s = c;
    for (j = 1; j <= sm; j++) {
      cout << z << " ";
      for (long long r = 1; r <= z; r++) {
        cout << s << " ";
        s++;
        if (s > n) s = 1;
      }
      cout << '\n';
    }
  }
  return;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long test = 2;
  switch (test) {
    case 1:
      solve();
      break;
    case 2: {
      long long t;
      cin >> t;
      while (t--) {
        solve();
      }
    }; break;
  }
}

### Prompt

Develop a solution in cpp to the problem described below:

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;
const long long MOD = pow(10, 9) + 7;
long long fac[1001];
template <typename T, typename U>
static inline void amin(T &x, U y) {
  if (y < x) x = y;
}
template <typename T, typename U>
static inline void amax(T &x, U y) {
  if (x < y) x = y;
}
void SieveOfEratosthenes(long long n) {
  bool prime[n + 1];
  memset(prime, true, sizeof(prime));
  for (long long p = 2; p * p <= n; p++) {
    if (prime[p] == true) {
      for (long long i = p * p; i <= n; i += p) prime[i] = false;
    }
  }
  for (long long p = 2; p <= n; p++)
    if (prime[p]) {
    }
}
long long power(long long x, long long y, long long p) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long modInverse(long long n, long long p) { return power(n, p - 2, p); }
long long nCrModPFermat(long long n, long long r, long long p) {
  if (n < r) return 0;
  if (r == 0) return 1;
  return (fac[n] * modInverse(fac[r], p) % p * modInverse(fac[n - r], p) % p) %
         p;
}
void dfs(long long v, vector<vector<long long>> &adj,
         vector<long long> &visited) {
  for (long long u : adj[v]) {
    if (!visited[u]) {
      visited[u] = 1;
    }
  }
}
long long parent[300002], sz[300002];
long long find_set(long long v) {
  if (v == parent[v]) return v;
  return parent[v] = find_set(parent[v]);
}
void make_set(long long v) {
  parent[v] = v;
  sz[v] = 1;
}
void union_sets(long long a, long long b) {
  a = find_set(a);
  b = find_set(b);
  if (a != b) {
    if (sz[a] < sz[b]) swap(a, b);
    parent[b] = a;
    sz[a] += sz[b];
  }
}
class segTree {
 public:
  long long len;
  vector<long long> a;
  vector<pair<long long, long long>> t;
  segTree(vector<long long> &val) {
    len = val.size();
    a.resize(len);
    long long i;
    a = val;
    if ((len & (len - 1)) != 0) {
      long long np = ceil(log2(len));
      long long pval = pow(2, np);
      long long diff = pval - len;
      len = pval;
      for (i = 0; i < diff; i++) a.push_back(pow(10, 10));
    }
    t.resize(4 * len + 1);
  }
  void build(long long idx, long long tl, long long tr) {
    if (tl == tr) {
      t[idx] = {a[tl], 1};
      return;
    }
    long long tm = (tl + tr) / 2;
    build(idx * 2, tl, tm);
    build(idx * 2 + 1, tm + 1, tr);
    long long count = 0;
    long long mn = min(t[2 * idx].first, t[2 * idx + 1].first);
    if (mn == t[2 * idx].first) count += t[2 * idx].second;
    if (mn == t[2 * idx + 1].first) count += t[2 * idx + 1].second;
    t[idx] = {mn, count};
  }
  pair<long long, long long> query(long long idx, long long tl, long long tr,
                                   long long l, long long r) {
    if (l > r) return {pow(10, 10), 1};
    if (l == tl && r == tr) {
      return t[idx];
    }
    long long tm = (tl + tr) / 2;
    pair<long long, long long> res1 = query(idx * 2, tl, tm, l, min(r, tm));
    pair<long long, long long> res2 =
        query(idx * 2 + 1, tm + 1, tr, max(l, tm + 1), r);
    if (res1.first < res2.first) {
      return res1;
    } else if (res2.first < res1.first) {
      return res2;
    } else {
      return {res1.first, res1.second + res2.second};
    }
  }
  void update(long long idx, long long tl, long long tr, long long pos,
              long long val) {
    if (tl == tr) {
      t[idx] = {val, 1};
      return;
    }
    long long tm = (tl + tr) / 2;
    if (pos <= tm)
      update(2 * idx, tl, tm, pos, val);
    else
      update(2 * idx + 1, tm + 1, tr, pos, val);
    long long count = 0;
    long long mn = min(t[2 * idx].first, t[2 * idx + 1].first);
    if (mn == t[2 * idx].first) count += t[2 * idx].second;
    if (mn == t[2 * idx + 1].first) count += t[2 * idx + 1].second;
    t[idx] = {mn, count};
  }
};
void solve() {
  long long i, j, n, m, k;
  cin >> n >> m >> k;
  if (n % m == 0) {
    for (i = 1; i <= k; i++) {
      for (j = 1; j <= m; j++) {
        long long z = n / m;
        cout << z << " ";
        long long s = z * (j - 1) + 1;
        long long e = z * j;
        for (int r = s; r <= e; r++) {
          cout << r << " ";
        }
        cout << '\n';
      }
    }
    return;
  }
  long long big = n % m, sm = m - big;
  long long c = 1;
  for (i = 1; i <= k; i++) {
    long long z = n / m;
    for (j = 1; j <= big; j++) {
      cout << z + 1 << " ";
      for (long long r = 1; r <= z + 1; r++) {
        cout << c << " ";
        c++;
        if (c > n) c = 1;
      }
      cout << '\n';
    }
    long long s = c;
    for (j = 1; j <= sm; j++) {
      cout << z << " ";
      for (long long r = 1; r <= z; r++) {
        cout << s << " ";
        s++;
        if (s > n) s = 1;
      }
      cout << '\n';
    }
  }
  return;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long test = 2;
  switch (test) {
    case 1:
      solve();
      break;
    case 2: {
      long long t;
      cin >> t;
      while (t--) {
        solve();
      }
    }; break;
  }
}

```