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: /*
  stream Butter!
  eggyHide eggyVengeance
  I need U
  xiao rerun when
   */
  import static java.lang.Math.*;
  import static java.lang.System.out;
  import java.util.*;
  import java.io.*;
  import java.math.*;
  
  public class x1552E2
  {
      public static void main(String hi[]) throws Exception
      {
          BufferedReader infile = new BufferedReader(new InputStreamReader(System.in));
          StringTokenizer st = new StringTokenizer(infile.readLine());
          int N = Integer.parseInt(st.nextToken());
          int K = Integer.parseInt(st.nextToken());
          int[] arr = readArr(N*K, infile, st);
          for(int i=0; i < N*K; i++)
              arr[i]--;
          int[] freq = new int[N*K];
          int[][] res = new int[N][2];
          int[] active = new int[N];
          int BOUND = (N+K-2)/(K-1);
          for(int i=0; i < N*K; i++)
          {
              if(active[arr[i]] == 0)
                  active[arr[i]] = i+1;
              else if(res[arr[i]][0] == 0)
              {
                  res[arr[i]][0] = active[arr[i]];
                  res[arr[i]][1] = i+1;
                  int max = 0;
                  for(int a=active[arr[i]]-1; a <= i; a++)
                  {
                      freq[a]++;
                      max = max(max, freq[a]);
                  }
                  if(max == BOUND)
                      Arrays.fill(active, 0);
              }
          }
          StringBuilder sb = new StringBuilder();
          for(int c=0; c < N; c++)
              sb.append(res[c][0]+" "+res[c][1]+"\n");
          System.out.print(sb);
      }
      public static int[] readArr(int N, BufferedReader infile, StringTokenizer st) throws Exception
      {
          int[] arr = new int[N];
          st = new StringTokenizer(infile.readLine());
          for(int i=0; i < N; i++)
              arr[i] = Integer.parseInt(st.nextToken());
          return arr;
      }
  }
  /*
  ai and bi of chosen intervals must be consecutive occurrences of that color
  
  3 4
  1 3 2 3 2 1 3 1 3 2 1 2
  
  3 4
  1 3 2 3 2 1 3 1 2 1 2 3
  
  6 3
  1 2 3 4 5 6 1 2 3 4 5 6 1 2 3 4 5 6
   */