Problem: Recently Pashmak has been employed in a transportation company. The company has k buses and has a contract with a school which has n students. The school planned to take the students to d different places for d days (each day in one place). Each day the company provides all the buses for the trip. Pashmak has to arrange the students in the buses. He wants to arrange the students in a way that no two students become close friends. In his ridiculous idea, two students will become close friends if and only if they are in the same buses for all d days.

Please help Pashmak with his weird idea. Assume that each bus has an unlimited capacity.

Input

The first line of input contains three space-separated integers n, k, d (1 ≤ n, d ≤ 1000; 1 ≤ k ≤ 109).

Output

If there is no valid arrangement just print -1. Otherwise print d lines, in each of them print n integers. The j-th integer of the i-th line shows which bus the j-th student has to take on the i-th day. You can assume that the buses are numbered from 1 to k.

Examples

Input

3 2 2


Output

1 1 2 
1 2 1 


Input

3 2 1


Output

-1

Note

Note that two students become close friends only if they share a bus each day. But the bus they share can differ from day to day.

Solution (in JAVA):

Solution:
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
	public static void main (String[] args) {
		InputStream inputStream = System.in;
		OutputStream outputStream = System.out;
		InputReader in = new InputReader (inputStream);
		PrintWriter out = new PrintWriter (outputStream);
		TaskC solver = new TaskC ();
		solver.solve (1, in, out);
		out.close ();
	}
	
	static class TaskC {
		public void solve (int testNumber, InputReader in, PrintWriter out) {
			int n = in.nextInt ();
			int k = in.nextInt ();
			int d = in.nextInt ();
//we are going to use pigeonhole principal and unique permutation generator here
//according to pegionhole if a student can take k buses at one day, then there
//can be k^d combinations in d days so here hole is k^d
			if (n>Math.round (Math.pow (k, d))) {
				out.println (-1);
				return;
			}
//now we are going to apply unique permuation or number generator in a range
// in this technique we will have to know about the range of numbers (suppose
//range is 2-4 then actually there are 3 numbers in range so basically we
// weill create such permutations where the range is between 3
//when we know the range of a number we can call it base of that number, so basically
//we are going to create 3 based numbers' unique permutation, five numbers of base 3
//can be 00 01 02 10 11, now as the range of numbers doesn't start with 0 so we will
//add 2 with every digit: 22 23 24 32 33, see this is the unique permutation
//as we all know base changing in hsc way is very easy and can be done to every base
			int[] os = new int[n+5];
			int j, res;
			int[][] ans = new int[d+5][n+5];
			for (int i = 0; i<n; i++) {
				res = i;
				j = 0;
				while (res != 0) {
					os[j++] = res%k;
					res /= k;
				}
				int cnt = j;
				j = 0;
				for (int l = d-1; l>=0; l--) {
					ans[l][i] = j<cnt ? os[j++] : 0;
				}
			}
			for (int i = 0; i<d; i++) {
				for (int l = 0; l<n; l++) {
					out.print ((ans[i][l]+1)+" ");
				}
				out.println ();
			}
		}
		
	}
	
	static class InputReader {
		private BufferedReader reader;
		private StringTokenizer tokenizer;
		
		public InputReader (InputStream stream) {
			reader = new BufferedReader (new InputStreamReader (stream), 32768);
			tokenizer = null;
		}
		
		public String next () {
			while (tokenizer == null || !tokenizer.hasMoreTokens ()) {
				try {
					String str;
					if ((str = reader.readLine ()) != null) tokenizer = new StringTokenizer (str);
					else return null;//to detect eof
				} catch (IOException e) {
					throw new RuntimeException (e);
				}
			}
			return tokenizer.nextToken ();
		}
		
		public int nextInt () {
			return Integer.parseInt (next ());
		}
		
	}
}