description: 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
solution: import java.io.BufferedReader;
  import java.io.IOException;
  import java.io.InputStreamReader;
  import java.io.PrintWriter;
  import java.text.DecimalFormat;
  import java.util.*;
  
   
  public class Codeforces {
  	static int mod= 1000000007;
  	
  	public static void main(String[] args) throws Exception {
  		PrintWriter out=new PrintWriter(System.out);
  		FastScanner fs=new FastScanner();
  		int t=fs.nextInt();
  		
  		outer:while(t-->0) {
  			int n=fs.nextInt(),m=fs.nextInt(), k=fs.nextInt();
  			int npb=(n+m-1)/m;
  			int npc=n/m;
  			int tb=n%m;
  			int incr= tb*npb;
  			int ans[][]=new int[m][];
  			int ind=0;
  			while(ind<tb) ans[ind++]=new int[npb];
  			while(ind<m) ans[ind++]=new int[npc];
  			ind=0;
  			for(int g=0;g<k;g++) {
  				for(int i=0;i<ans.length;i++){
  					for(int j=0;j<ans[i].length;j++) {
  						ans[i][j]=(ind+1);
  						ind++;
  						ind%=n;
  					}
  				}
  				for(int i=0;i<ans.length;i++) {
  					out.print(ans[i].length+" ");
  					for(int j=0;j<ans[i].length;j++) {
  						out.print(ans[i][j]+" ");
  					}
  					out.println();
  				}
  				ind=(ind+incr)%n;
  			}
  			out.println();
  		}
  		
  		out.close();
  		
  	}
  	
  	static long pow(long a,long b) {
  		if(b<0) return 1;
  		long res=1;
  		while(b!=0) {
  			if((b&1)!=0) {
  				res*=a;
  				res%=mod;
  			}
  			a*=a;
  			a%=mod;
  			b=b>>1;
  		}
  		return res;
  	}
  	static int gcd(int  a,int  b) {
  		if(b==0) return a;
  		return gcd(b,a%b);
  	}
  	static long nck(int n,int k) {
  		if(k>n) return 0;
  		long res=1;
  		res*=fact(n);
  		res%=mod;
  		res*=modInv(fact(k));
  		res%=mod;
  		res*=modInv(fact(n-k)); 
  		res%=mod;
  		return res;
  	}
  	static long fact(long n) {
  //		return fact[(int)n];
  		long res=1;
  		for(int i=2;i<=n;i++) {
  			res*=i;
  			res%=mod;
  		}
  		return res;
  	}
  	
  	static long modInv(long n) {
  		return pow(n,mod-2);
  	}
  	
  	static void sort(int[] a) {
  		//suffle
  		int n=a.length;
  		Random r=new Random();
  		for (int i=0; i<a.length; i++) {
  			int oi=r.nextInt(n);
  			int temp=a[i];
  			a[i]=a[oi];
  			a[oi]=temp;
  		}
  		
  		//then sort
  		Arrays.sort(a);
  	}
  	
  	// Use this to input code since it is faster than a Scanner
  	static class FastScanner {
  		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
  		StringTokenizer st=new StringTokenizer("");
  		String next() {
  			while (!st.hasMoreTokens())
  				try {
  					st=new StringTokenizer(br.readLine());
  				} catch (IOException e) {
  					e.printStackTrace();
  				}
  			return st.nextToken();
  		}
  		
  		int nextInt() {
  			return Integer.parseInt(next());
  		}
  		long[] lreadArray(int n) {
  			long a[]=new long[n];
  			for(int i=0;i<n;i++) a[i]=nextLong();
  			return a;
  		}
  		int[] readArray(int n) {
  			int[] a=new int[n];
  			for (int i=0; i<n; i++) a[i]=nextInt();
  			return a;
  		}
  		long nextLong() {
  			return Long.parseLong(next());
  		}
  	}
   
  }