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.util.*;
import java.io.*;
import java.math.*;
import java.lang.*;
import java.util.stream.*;

/**
 * @author Pradyumn
 */

public class Main implements Runnable{
	public int TC = 0;
	public static void main(String[] args) {
		new Thread(null, new Main(), ":)", 1L << 26).start();
	}
	@SuppressWarnings("Unchecked")
	public void solve() throws IOException {
		int n = ni(), K = ni(), d = ni();
		long kk = 1;
		for(int i = 0;i < d;i++){
			kk *= K;
			if(kk >= n)kk = n+1;
		}
		if(kk < n){
			out.println(-1);
		}else{
			StringBuilder[] sbs = new StringBuilder[d];
			for(int i = 0;i < d;i++)sbs[i] = new StringBuilder();
			int[] num = new int[d];
			for(int i = 0;i < n;i++){
				for(int j = 0;j < d;j++){
					sbs[j].append(" " + (num[j]+1));
				}
				
				for(int j = d-1;j >= 0;j--){
					if(++num[j] >= K){
						num[j] = 0;
					}else{
						break;
					}
				}
			}
			for(int i = 0;i < d;i++){
				out.println(sbs[i].substring(1));
			}
		}
	}
	BufferedReader br;
	StringTokenizer st;
	PrintWriter out;
	public void run() {
		try {
			br = new BufferedReader(new InputStreamReader(System.in),32768);
			st = null;
			out = new PrintWriter(System.out);
			if(TC == 1) {
				for(int T = ni();T > 0; --T) {
					solve();
				}
			} else {
				solve();
			}
			br.close();
			out.close();
		} catch(IOException e) {
			e.printStackTrace();
			System.exit(1);
		}
	}
	public char[] ns(int n) throws IOException {
		return ns().toCharArray();
	}
	public char[][] nm(int n, int m) throws IOException {
		char[][] map = new char[n][];
		for(int i = 0;i < n; ++i) {
			map[i] = ns(m);
		}
		return map;
	}
	public String ns() throws IOException {
		while (st == null || !st.hasMoreTokens()) {
			st = new StringTokenizer(br.readLine());
		}
		return st.nextToken();
	}
	public String nextLine() throws IOException {
		return br.readLine();
	}
	public int ni() throws IOException {
		return Integer.parseInt(ns());
	}

	public int[] na(int n) throws IOException {
		int[] a = new int[n];
		for(int i = 0;i < n; ++i) a[i] = ni();
		return a;
	}

	public long nl() throws IOException {
		return Long.parseLong(ns());
	}

	public long[] nal(int n) throws IOException {
		long[] a = new long[n];
		for(int i = 0;i < n; ++i) a[i] = nl();
		return a;
	}

	public double nd() throws IOException {
		return Double.parseDouble(ns());
	}
	private static boolean oj = System.getProperty("ONLINE_JUDGE") != null;
	private static void tr(Object... o) { if(!oj)System.out.println(Arrays.deepToString(o)); } 
}