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 CF1552E extends PrintWriter {
  	CF1552E() { super(System.out); }
  	Scanner sc = new Scanner(System.in);
  	public static void main(String[] $) {
  		CF1552E o = new CF1552E(); o.main(); o.flush();
  	}
  
  	void main() {
  		int n = sc.nextInt();
  		int k = sc.nextInt();
  		int l = (n + k - 2) / (k - 1);
  		int[] aa = new int[n];
  		int[] bb = new int[n];
  		int[] ii = new int[n];
  		int[] hh = new int[n];
  		int[] ll = new int[k];
  		boolean[] used = new boolean[n];
  		for (int i = 0; i < n * k; i++) {
  			int c = sc.nextInt() - 1;
  			if (used[c])
  				continue;
  			int h = hh[c]++;
  			if (h > 0 && ll[h] < l) {
  				used[c] = true;
  				ll[h]++;
  				aa[c] = ii[c];
  				bb[c] = i;
  			} else
  				ii[c] = i;
  		}
  		for (int c = 0; c < n; c++)
  			println((aa[c] + 1) + " " + (bb[c] + 1));
  	}
  }
  