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.util.Scanner;
  
  public class E {
      public static void main(String[] args) {
          Scanner s = new Scanner(System.in);
  
          solve(s.nextInt(), s.nextInt(), s);
      }
  
  
      public static void solve(int c, int r, Scanner s) {
          int v = c / (r - 1) + (c % (r - 1) == 0 ? 0 : 1);
          int e = c / v + (c % v == 0 ? 0 : 1);
  
          int a[] = new int[c * r];
          for (int i = 0; i < a.length; i++) {
              a[i] = s.nextInt() - 1;
          }
          
          boolean done[];
          boolean visited[];
          int first[];
          int l;
          int min;
          int index;
          int t;
  
          String ans[] = new String[c];
  
          for (int i = 0; i < v; i++) {
              min = i * e;
              if (i != v - 1) {
                  l = e;
              } else {
                  l = c % e;
                  if (l == 0) l = e;
              }
  
              done = new boolean[l];
              index = 0;
              for (int j = 0; j < l; j++) {
                  first = new int[l];
                  visited = new boolean[l];
                  for (; index < a.length; index++) {
                      t = a[index];
                      if (t >= min && t < min + l) {
                          if (!done[t - min]) {
                              if (visited[t - min]) {
                                  ans[t] = (first[t - min] + 1) + " " + (index + 1);
                                  done[t - min] = true;
                                  break;
                              } else {
                                  visited[t - min] = true;
                                  first[t - min] = index;
                              }
                          }
                      }
                  }
              }
          }
  
          for (int i = 0; i < c; i++) {
              System.out.println(ans[i]);
          }
      }
  }
  