#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
using namespace std;
long long int fac[1000005];
long long int modularExponentiation(long long int x, long long int n,
                                    long long int M) {
  long long int result = 1;
  while (n > 0) {
    if (n % 2 == 1)
      result = ((result % 998244353ll) * (x % 998244353ll)) % 998244353ll;
    x = ((x % 998244353ll) * (x % 998244353ll)) % 998244353ll;
    n = n / 2;
  }
  return result;
}
long long int binaryExponentiation(long long int x, long long int n) {
  long long int result = 1.0;
  while (n > 0) {
    if (n % 2 == 1) result = result * x;
    x = x * x;
    n = n / 2;
  }
  return result;
}
long long int pow1(long long int x, long long int y) {
  long long int z = 1;
  while (y--) {
    z = z * x;
  }
  return z;
}
bool isprime(long long int n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (long long int i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
long long int modInverse(long long int n, long long int p) {
  return modularExponentiation(n, p - 2, p);
}
long long int nCrModPFermat(long long int n, long long int r, long long int p) {
  if (r == 0) return 1;
  return (fac[n] * modInverse(fac[r], p) % p * modInverse(fac[n - r], p) % p) %
         p;
}
long long int check(long long int x, long long int y) {
  long long int z = 1;
  long long int ans = 0;
  while (z < x) {
    ans++;
    z *= y;
  }
  return ans;
}
long long int countbits(long long int n) {
  long long int x = 0;
  while (n > 0) {
    x++;
    n = (n & (n - 1));
  }
  return x;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t, n, m, k;
  cin >> t;
  while (t--) {
    cin >> n >> m >> k;
    long long int r = n % m;
    if (r == 0) {
      long long int x = n / m;
      long long int p = 1;
      for (long long int j1 = 1; j1 <= k; j1++) {
        for (long long int j = 1; j <= m; j++) {
          cout << x << ' ';
          for (long long int z = 1; z <= x; z++) {
            cout << p << ' ';
            p++;
            if (p > n) {
              p = 1;
            }
          }
          cout << "\n";
        }
      }
    } else {
      long long int x = n / m;
      long long int p = 1;
      for (long long int j1 = 1; j1 <= k; j1++) {
        vector<long long int> v[m];
        for (long long int z = 1; z <= r; z++) {
          long long int T = x + 1;
          while (T--) {
            v[z - 1].push_back(p++);
            if (p > n) p = 1;
          }
        }
        long long int p1 = p;
        for (long long int j = r + 1; j <= m; j++) {
          for (long long int z = 1; z <= x; z++) {
            v[j - 1].push_back(p1++);
            if (p1 > n) p1 = 1;
          }
        }
        for (long long int j = 1; j <= m; j++) {
          cout << v[j - 1].size() << ' ';
          for (auto z : v[j - 1]) cout << z << ' ';
          cout << "\n";
        }
      }
    }
    cout << "\n";
  }
}

### 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>
#pragma GCC optimize("Ofast")
using namespace std;
long long int fac[1000005];
long long int modularExponentiation(long long int x, long long int n,
                                    long long int M) {
  long long int result = 1;
  while (n > 0) {
    if (n % 2 == 1)
      result = ((result % 998244353ll) * (x % 998244353ll)) % 998244353ll;
    x = ((x % 998244353ll) * (x % 998244353ll)) % 998244353ll;
    n = n / 2;
  }
  return result;
}
long long int binaryExponentiation(long long int x, long long int n) {
  long long int result = 1.0;
  while (n > 0) {
    if (n % 2 == 1) result = result * x;
    x = x * x;
    n = n / 2;
  }
  return result;
}
long long int pow1(long long int x, long long int y) {
  long long int z = 1;
  while (y--) {
    z = z * x;
  }
  return z;
}
bool isprime(long long int n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (long long int i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
long long int modInverse(long long int n, long long int p) {
  return modularExponentiation(n, p - 2, p);
}
long long int nCrModPFermat(long long int n, long long int r, long long int p) {
  if (r == 0) return 1;
  return (fac[n] * modInverse(fac[r], p) % p * modInverse(fac[n - r], p) % p) %
         p;
}
long long int check(long long int x, long long int y) {
  long long int z = 1;
  long long int ans = 0;
  while (z < x) {
    ans++;
    z *= y;
  }
  return ans;
}
long long int countbits(long long int n) {
  long long int x = 0;
  while (n > 0) {
    x++;
    n = (n & (n - 1));
  }
  return x;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t, n, m, k;
  cin >> t;
  while (t--) {
    cin >> n >> m >> k;
    long long int r = n % m;
    if (r == 0) {
      long long int x = n / m;
      long long int p = 1;
      for (long long int j1 = 1; j1 <= k; j1++) {
        for (long long int j = 1; j <= m; j++) {
          cout << x << ' ';
          for (long long int z = 1; z <= x; z++) {
            cout << p << ' ';
            p++;
            if (p > n) {
              p = 1;
            }
          }
          cout << "\n";
        }
      }
    } else {
      long long int x = n / m;
      long long int p = 1;
      for (long long int j1 = 1; j1 <= k; j1++) {
        vector<long long int> v[m];
        for (long long int z = 1; z <= r; z++) {
          long long int T = x + 1;
          while (T--) {
            v[z - 1].push_back(p++);
            if (p > n) p = 1;
          }
        }
        long long int p1 = p;
        for (long long int j = r + 1; j <= m; j++) {
          for (long long int z = 1; z <= x; z++) {
            v[j - 1].push_back(p1++);
            if (p1 > n) p1 = 1;
          }
        }
        for (long long int j = 1; j <= m; j++) {
          cout << v[j - 1].size() << ' ';
          for (auto z : v[j - 1]) cout << z << ' ';
          cout << "\n";
        }
      }
    }
    cout << "\n";
  }
}

```