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.*;
  import java.io.*;
  import java.text.*;
  
  public class CF_1552_E{
      //SOLUTION BEGIN
      void pre() throws Exception{}
      void solve(int TC) throws Exception{
          int N = ni(), K = ni();
          int[] C = new int[N*K];
          for(int i = 0; i< N*K; i++)C[i] = ni()-1;
  //        List<Integer>[] adj = new ArrayList[N];
          int[][] seg = new int[N][];
  
  
          int[] prev = new int[N];
          for(int i = 0; i< N; i += K-1){
              int lo = i, hi = Math.min(N-1, i+K-2);
              List<Integer> adj = new ArrayList<>();
              for(int j = 0; j < N*K; j++)if(lo <= C[j] && C[j] <= hi)adj.add(j);
  
              for(int j = lo; j <= hi; j++)prev[j] = -1;
              for(int p:adj){
                  if(prev[C[p]] != -1 && seg[C[p]] == null){
                      seg[C[p]] = new int[]{prev[C[p]], p};
                      for(int j = lo; j <= hi; j++)prev[j] = -1;
                  }else prev[C[p]] = p;
              }
  
          }
  
          for(int i = 0; i< N; i++)pn(1+seg[i][0], 1+seg[i][1]);
          int[] sum = new int[1+N*K];
          for(int i = 0; i< N; i++){
              sum[seg[i][0]]++;
              sum[seg[i][1]+1]--;
          }
          for(int i = 0; i< sum.length-1; i++)sum[i+1] += sum[i];
          for(int x:sum)hold(x <= (N+K-2)/(K-1));
      }
      //SOLUTION END
      void hold(boolean b)throws Exception{if(!b)throw new Exception("Hold right there, Sparky!");}
      void exit(boolean b){if(!b)System.exit(0);}
      static void dbg(Object... o){System.err.println(Arrays.deepToString(o));}
      final long IINF = (long)1e17;
      final int INF = (int)1e9+2;
      DecimalFormat df = new DecimalFormat("0.00000000000");
      double PI = 3.141592653589793238462643383279502884197169399, eps = 1e-8;
      static boolean multipleTC = false, memory = true, fileIO = false;
      FastReader in;PrintWriter out;
      void run() throws Exception{
          long ct = System.currentTimeMillis();
          if (fileIO) {
              in = new FastReader("");
              out = new PrintWriter("");
          } else {
              in = new FastReader();
              out = new PrintWriter(System.out);
          }
          //Solution Credits: Taranpreet Singh
          int T = multipleTC? ni():1;
          pre();
          for (int t = 1; t <= T; t++) solve(t);
          out.flush();
          out.close();
          System.err.println(System.currentTimeMillis() - ct);
      }
      public static void main(String[] args) throws Exception{
          if(memory)new Thread(null, new Runnable() {public void run(){try{new CF_1552_E().run();}catch(Exception e){e.printStackTrace();System.exit(1);}}}, "1", 1 << 28).start();
          else new CF_1552_E().run();
      }
      int[][] make(int n, int e, int[] from, int[] to, boolean f){
          int[][] g = new int[n][];int[]cnt = new int[n];
          for(int i = 0; i< e; i++){
              cnt[from[i]]++;
              if(f)cnt[to[i]]++;
          }
          for(int i = 0; i< n; i++)g[i] = new int[cnt[i]];
          for(int i = 0; i< e; i++){
              g[from[i]][--cnt[from[i]]] = to[i];
              if(f)g[to[i]][--cnt[to[i]]] = from[i];
          }
          return g;
      }
      int[][][] makeS(int n, int e, int[] from, int[] to, boolean f){
          int[][][] g = new int[n][][];int[]cnt = new int[n];
          for(int i = 0; i< e; i++){
              cnt[from[i]]++;
              if(f)cnt[to[i]]++;
          }
          for(int i = 0; i< n; i++)g[i] = new int[cnt[i]][];
          for(int i = 0; i< e; i++){
              g[from[i]][--cnt[from[i]]] = new int[]{to[i], i, 0};
              if(f)g[to[i]][--cnt[to[i]]] = new int[]{from[i], i, 1};
          }
          return g;
      }
      int find(int[] set, int u){return set[u] = (set[u] == u?u:find(set, set[u]));}
      int digit(long s){int ans = 0;while(s>0){s/=10;ans++;}return ans;}
      long gcd(long a, long b){return (b==0)?a:gcd(b,a%b);}
      int gcd(int a, int b){return (b==0)?a:gcd(b,a%b);}
      int bit(long n){return (n==0)?0:(1+bit(n&(n-1)));}
      void p(Object... o){for(Object oo:o)out.print(oo+" ");}
      void pn(Object... o){for(int i = 0; i< o.length; i++)out.print(o[i]+(i+1 < o.length?" ":"\n"));}
      void pni(Object... o){for(Object oo:o)out.print(oo+" ");out.println();out.flush();}
      String n()throws Exception{return in.next();}
      String nln()throws Exception{return in.nextLine();}
      int ni()throws Exception{return Integer.parseInt(in.next());}
      long nl()throws Exception{return Long.parseLong(in.next());}
      double nd()throws Exception{return Double.parseDouble(in.next());}
  
      class FastReader{
          BufferedReader br;
          StringTokenizer st;
          public FastReader(){
              br = new BufferedReader(new InputStreamReader(System.in));
          }
  
          public FastReader(String s) throws Exception{
              br = new BufferedReader(new FileReader(s));
          }
  
          String next() throws Exception{
              while (st == null || !st.hasMoreElements()){
                  try{
                      st = new StringTokenizer(br.readLine());
                  }catch (IOException  e){
                      throw new Exception(e.toString());
                  }
              }
              return st.nextToken();
          }
  
          String nextLine() throws Exception{
              String str;
              try{
                  str = br.readLine();
              }catch (IOException e){
                  throw new Exception(e.toString());
              }
              return str;
          }
      }
  }