#include <bits/stdc++.h>
using namespace std;
const int maxdouble = 0x7f;
const int mindouble = 0xfe;
const long long mod = 998244353;
double jz(int n) {
  double ans = 1;
  for (int i = 1; i <= n; i++) {
    ans *= i;
  }
  return ans;
}
inline long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch > '9' || ch < '0') {
    if (ch == '-') f = -1;
    ch = getchar();
    if (ch == -1) return 0;
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
inline void write(long long x) {
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  if (x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}
long long gcd(long long a, long long b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
inline long long check(long long x) {
  long long s = sqrt(x * 2) + (1e-12);
  if ((s * (s + 1) / 2) == x) return s;
  return -1;
}
long long int ksm(long long int a, long long int b) {
  long long int ret = 1;
  while (b) {
    if (b & 1) ret = ret * a % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return ret;
}
inline long long mul(long long a, long long b, long long mod) {
  return (a * b - (long long)((long double)a / mod * b) * mod + mod) % mod;
}
int main(void) {
  int T;
  scanf("%d", &T);
  while (T--) {
    int n, m, k;
    scanf("%d%d%d", &n, &m, &k);
    int l = 0, a = n / m, b = (n + m - 1) / m, t2 = n % m, t1 = m - t2;
    while (k--) {
      int t = l;
      for (int i = 1; i <= t2; i++) {
        printf("%d", b);
        for (int j = 0; j < b; j++) {
          printf(" %d", (t + j) % n + 1);
        }
        cout << '\n';
        ;
        t = (t + b) % n;
      }
      for (int i = 1; i <= t1; i++) {
        printf("%d", a);
        for (int j = 0; j < a; j++) {
          printf(" %d", (t + j) % n + 1);
        }
        cout << '\n';
        ;
        t = (t + a) % n;
      }
      l = (l + b * t2) % n;
    }
    cout << '\n';
    ;
  }
  return 0;
}

### Prompt

Please provide a cpp coded solution 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 int maxdouble = 0x7f;
const int mindouble = 0xfe;
const long long mod = 998244353;
double jz(int n) {
  double ans = 1;
  for (int i = 1; i <= n; i++) {
    ans *= i;
  }
  return ans;
}
inline long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch > '9' || ch < '0') {
    if (ch == '-') f = -1;
    ch = getchar();
    if (ch == -1) return 0;
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
inline void write(long long x) {
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  if (x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}
long long gcd(long long a, long long b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
inline long long check(long long x) {
  long long s = sqrt(x * 2) + (1e-12);
  if ((s * (s + 1) / 2) == x) return s;
  return -1;
}
long long int ksm(long long int a, long long int b) {
  long long int ret = 1;
  while (b) {
    if (b & 1) ret = ret * a % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return ret;
}
inline long long mul(long long a, long long b, long long mod) {
  return (a * b - (long long)((long double)a / mod * b) * mod + mod) % mod;
}
int main(void) {
  int T;
  scanf("%d", &T);
  while (T--) {
    int n, m, k;
    scanf("%d%d%d", &n, &m, &k);
    int l = 0, a = n / m, b = (n + m - 1) / m, t2 = n % m, t1 = m - t2;
    while (k--) {
      int t = l;
      for (int i = 1; i <= t2; i++) {
        printf("%d", b);
        for (int j = 0; j < b; j++) {
          printf(" %d", (t + j) % n + 1);
        }
        cout << '\n';
        ;
        t = (t + b) % n;
      }
      for (int i = 1; i <= t1; i++) {
        printf("%d", a);
        for (int j = 0; j < a; j++) {
          printf(" %d", (t + j) % n + 1);
        }
        cout << '\n';
        ;
        t = (t + a) % n;
      }
      l = (l + b * t2) % n;
    }
    cout << '\n';
    ;
  }
  return 0;
}

```