description: The numbers 1,   2,   ...,   n ⋅ k are colored with n colors. These colors are indexed by 1,   2,   ...,   n. For each 1 ≤ i ≤ n, there are exactly k numbers colored with color i.
  
  Let [a,   b] denote the interval of integers between a and b inclusive, that is, the set \\{a,   a + 1,   ...,   b\}. You must choose n intervals [a_1,   b_1],   [a_2,   b_2],   ..., [a_n,   b_n] such that: 
  
    * for each 1 ≤ i ≤ n, it holds 1 ≤ a_i < b_i ≤ n ⋅ k; 
    * for each 1 ≤ i ≤ n, the numbers a_i and b_i are colored with color i; 
    * each number 1 ≤ x ≤ n ⋅ k belongs to at most \left⌈ (n)/(k - 1) \right⌉ intervals. 
  
  
  
  One can show that such a family of intervals always exists under the given constraints.
  
  Input
  
  The first line contains two integers n and k (1 ≤ n ≤ 100, 2 ≤ k ≤ 100) — the number of colors and the number of occurrences of each color.
  
  The second line contains n ⋅ k integers c_1,   c_2,   ...,   c_{nk} (1 ≤ c_j ≤ n), where c_j is the color of number j. It is guaranteed that, for each 1 ≤ i ≤ n, it holds c_j = i for exactly k distinct indices j.
  
  Output
  
  Output n lines. The i-th line should contain the two integers a_i and b_i.
  
  If there are multiple valid choices of the intervals, output any.
  
  Examples
  
  Input
  
  
  4 3
  2 4 3 1 1 4 2 3 2 1 3 4
  
  
  Output
  
  
  4 5
  1 7
  8 11
  6 12
  
  Input
  
  
  1 2
  1 1
  
  
  Output
  
  
  1 2
  
  
  Input
  
  
  3 3
  3 1 2 3 2 1 2 1 3
  
  
  Output
  
  
  6 8
  3 7
  1 4
  
  Input
  
  
  2 3
  2 1 1 1 2 2
  
  
  Output
  
  
  2 3
  5 6
  
  Note
  
  In the first sample, each number can be contained in at most \left⌈ (4)/(3 - 1) \right⌉ = 2 intervals. The output is described by the following picture:
  
  <image>
  
  In the second sample, the only interval to be chosen is forced to be [1,   2], and each number is indeed contained in at most \left⌈ (1)/(2 - 1) \right⌉ = 1 interval.
  
  In the third sample, each number can be contained in at most \left⌈ (3)/(3 - 1) \right⌉ = 2 intervals. The output is described by the following picture:
  
  <image>
solution: import java.io.*;
  import java.util.*;
  
  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);
  		Task solver = new Task();
  		int tcCnt = 1;
  		for (int tc = 1; tc <= tcCnt; tc++)
  			solver.solve(tc, in, out);
  		out.close();
  	}
  
  	static class Task {
  		public void solve(int testNumber, InputReader in, PrintWriter out) {
  			int n = in.nextInt();
  			int k = in.nextInt();
  			int m = (n+k-2)/(k-1);
  			int[][] pos = new int[n][k];
  			int[] ptr = new int[n];
  			for (int i = 0; i < n*k; i++) {
  				int c = in.nextInt() - 1;
  				pos[c][ptr[c]++] = i;
  			}
  			Integer[] colors = new Integer[n];
  			for (int i = 0; i < n; i++) {
  				colors[i] = i;
  			}
  			int[][] intervals = new int[n][2];
  			int cnt = 0;
  			for (int i = 0; cnt < n; i++) {
  				int idx = i;
  				Arrays.sort(colors, Comparator.comparingInt(p -> pos[p][idx + 1]));
  				for (int j = 0, tmpcnt = 0; j < n && tmpcnt < m; j++) {
  					int col = colors[j];
  					if (intervals[col][0] != 0) {
  						continue;
  					}
  					tmpcnt++;
  					cnt++;
  					intervals[col][0] = pos[col][i] + 1;
  					intervals[col][1] = pos[col][i+1] + 1;
  				}
  			}
  			for (int i = 0; i < n; i++) {
  				out.println(intervals[i][0]+" "+intervals[i][1]);
  			}
  		}
  
  	}
  
  	static class InputReader {
  		public BufferedReader reader;
  		public StringTokenizer tokenizer;
  
  		public InputReader(InputStream stream) {
  			reader = new BufferedReader(new InputStreamReader(stream), 32768);
  			tokenizer = null;
  		}
  
  		public String nextLine() {
  			try {
  				return reader.readLine();
  			} catch (IOException e) {
  				throw new RuntimeException(e);
  			}
  		}
  
  		public String nextToken() {
  			while (tokenizer == null || !tokenizer.hasMoreTokens()) {
  				tokenizer = new StringTokenizer(nextLine());
  			}
  			return tokenizer.nextToken();
  		}
  
  		public int nextInt() {
  			return Integer.parseInt(nextToken());
  		}
  
  		public double nextDouble() {
  			return Double.parseDouble(nextToken());
  		}
  
  		public long nextLong() {
  			return Long.parseLong(nextToken());
  		}
  
  		public int[] nextIntArray(int size) {
  			int[] arr = new int[size];
  			for (int i = 0; i < size; i++) {
  				arr[i] = nextInt();
  			}
  			return arr;
  		}
  
  	}
  }