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.OutputStream;
  import java.io.IOException;
  import java.io.InputStream;
  import java.io.PrintWriter;
  import java.util.Arrays;
  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
   *
   * @author lucasr
   */
  public class Main {
  	public static void main(String[] args) {
  		InputStream inputStream = System.in;
  		OutputStream outputStream = System.out;
  		MyScanner in = new MyScanner(inputStream);
  		PrintWriter out = new PrintWriter(outputStream);
  		EColorsAndIntervals solver = new EColorsAndIntervals();
  		solver.solve(1, in, out);
  		out.close();
  	}
  
  	static class EColorsAndIntervals {
  		public static MyScanner sc;
  		public static PrintWriter out;
  
  		public void solve(int testNumber, MyScanner sc, PrintWriter out) {
  			EColorsAndIntervals.sc = sc;
  			EColorsAndIntervals.out = out;
  			int n = sc.nextInt();
  			int k = sc.nextInt();
  			IntArray[] pos = new IntArray[n];
  			for (int i = 0; i < n; i++) {
  				pos[i] = new IntArray();
  			}
  			for (int i = 1; i <= n * k; i++) {
  				int color = sc.nextInt() - 1;
  				pos[color].add(i);
  			}
  			int from = 0;
  			int[] resA = new int[n], resB = new int[n];
  			boolean[] used = new boolean[n];
  			while (from < n) {
  				int to = Math.min(from + k - 1, n);
  				int ops = to - from;
  				for (int end = 1; end < k && end <= ops; end++) {
  					int min = -1;
  					for (int i = from; i < to; i++)
  						if (!used[i]) {
  							if (min == -1 || pos[i].get(end) < pos[min].get(end)) {
  								min = i;
  							}
  						}
  					resA[min] = pos[min].get(end - 1);
  					resB[min] = pos[min].get(end);
  					used[min] = true;
  				}
  				from += k - 1;
  			}
  			for (int i = 0; i < n; i++) {
  				out.println(resA[i] + " " + resB[i]);
  			}
  		}
  
  	}
  
  	static class IntArray {
  		int[] arr;
  		int size;
  
  		public IntArray() {
  			arr = new int[4];
  		}
  
  		public void add(int val) {
  			if (size == arr.length) {
  				arr = Arrays.copyOf(arr, 2 * arr.length);
  			}
  			arr[size++] = val;
  		}
  
  		public int get(int pos) {
  			return arr[pos];
  		}
  
  		public int[] toArray() {
  			return Arrays.copyOf(arr, size);
  		}
  
  		public String toString() {
  			return "IntArray " + Arrays.toString(toArray());
  		}
  
  	}
  
  	static class MyScanner {
  		private BufferedReader br;
  		private StringTokenizer tokenizer;
  
  		public MyScanner(InputStream is) {
  			br = new BufferedReader(new InputStreamReader(is));
  		}
  
  		public String next() {
  			while (tokenizer == null || !tokenizer.hasMoreTokens()) {
  				try {
  					tokenizer = new StringTokenizer(br.readLine());
  				} catch (IOException e) {
  					throw new RuntimeException(e);
  				}
  			}
  			return tokenizer.nextToken();
  		}
  
  		public int nextInt() {
  			return Integer.parseInt(next());
  		}
  
  	}
  }
  
  