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.ArrayList;
  import java.util.Arrays;
  import java.util.StringTokenizer;
  import java.util.List;
   import java.util.*;
   public class realfast implements Runnable 
    {
      private static final int INF = (int) 1e9;
      long in= 1000000007;
      long fac[]= new long[1000001];
      long inv[]=new long[1000001];
      public  void solve() throws IOException 
      {
  
          
           
  
          int n = readInt();
          int k = readInt();
          int arr[]=new int[n*k+1];
  
          for(int i=1;i<=(n*k);i++)
            arr[i]= readInt();
          int map1[]=new int[n+1];
          Arrays.fill(map1,n*k+100);
          int next[]=new int[n*k+1];
          int s[]=new int[n+1];
          int e[]=new int[n+1];
          for(int i=n*k;i>=1;i--)
          {
                next[i]= map1[arr[i]];
                e[arr[i]]= next[i];
                s[arr[i]]=i;
                map1[arr[i]]=i;
          }
          int ans[][]=new int[n+1][2];
          int pal[]=new int[n+1];
          int g = (n+k-2)/(k-1);
         // out.println(g);
          boolean map[]=new boolean[n+1];
          for(int j=1;j<=n;j++)
          {
            int min = n*k+100;
            int cur=0;
            for(int i=1;i<=n;i++)
            {
              if(!map[i])
              {
                 if(e[i]<min)
                 {
                     min=e[i];
                     cur=i;
                 }
              }
            }
            ans[cur][0]=s[cur];
            ans[cur][1]=e[cur];
            map[cur]= true;
            for(int i=1;i<=n;i++)
            {
               if(!map[i])
               {
                if(e[i]<s[cur]||e[cur]<s[i])
                {
                  
                }
                else{
                  pal[i]++;
                 if(pal[i]==g)
                 {
                  pal[i]=0;
                  s[i]=e[i];
                  e[i]= next[e[i]];
                  }
                }
               }
  
            }        
          }
  
          for(int i=1;i<=n;i++)
          {
            out.println(s[i]+" "+e[i]);
          }
  
  
  
      }
      public boolean check(int arr[] , int i , int val, int n)
      {
          if(i==n)
          {
              if(Math.abs(val)==Math.abs(arr[n]))
                  return true;
              else
                  return false;
          }
          if(check(arr,i+1,val+arr[i],n)||check(arr,i+1,val-arr[i],n)||check(arr,i+1,val,n))
          {
              return true;
          }
          else
              return false;
      }
      
      public int value (int seg[], int left , int right ,int index, int l, int r)
      {
              
              if(left>right)
              {
                return -100000000;
              }
              if(right<l||left>r)
                  return -100000000;
              if(left>=l&&right<=r)
                  return seg[index];
              int mid = left+(right-left)/2;
              int val = value(seg,left,mid,2*index+1,l,r);
              int val2 = value(seg,mid+1,right,2*index+2,l,r);
              return Math.max(val,val2);
  
      }
     
      public int gcd(int a , int b )
      {
        if(a<b)
        {
          int t =a;
          a=b;
          b=t;
        }
        if(a%b==0)
          return b ;
        return gcd(b,a%b);
      }
      public long pow(long n , long p,long m)
      {
           if(p==0)
              return 1;
          long val = pow(n,p/2,m);;
          val= (val*val)%m;
          if(p%2==0)
              return val;
          else
              return (val*n)%m;
      }
      
      
      ///////////////////////////////////////////////////////////////////////////////////////////////////////////////////
      public static void main(String[] args) {
          new Thread(null, new realfast(), "", 128 * (1L << 20)).start();
      }
   
      private static final boolean ONLINE_JUDGE = System.getProperty("ONLINE_JUDGE") != null;
      private BufferedReader reader;
      private StringTokenizer tokenizer;
      private PrintWriter out;
   
      @Override
      public void run() {
          try {
              if (ONLINE_JUDGE || !new File("input.txt").exists()) {
                  reader = new BufferedReader(new InputStreamReader(System.in));
                  out = new PrintWriter(System.out);
              } else {
                  reader = new BufferedReader(new FileReader("input.txt"));
                  out = new PrintWriter("output.txt");
              }
              solve();
          } catch (IOException e) {
              throw new RuntimeException(e);
          } finally {
              try {
                  reader.close();
              } catch (IOException e) {
                  // nothing
              }
              out.close();
          }
      }
   
      private String readString() throws IOException {
          while (tokenizer == null || !tokenizer.hasMoreTokens()) {
              tokenizer = new StringTokenizer(reader.readLine());
          }
          return tokenizer.nextToken();
      }
   
      @SuppressWarnings("unused")
      private int readInt() throws IOException {
          return Integer.parseInt(readString());
      }
   
      @SuppressWarnings("unused")
      private long readLong() throws IOException {
          return Long.parseLong(readString());
      }
   
      @SuppressWarnings("unused")
      private double readDouble() throws IOException {
          return Double.parseDouble(readString());
      }
  }
  class edge implements Comparable<edge>{
      long u ;
      long v;
      
      edge(long u, long v)
      {
         this.u=u;
         this.v=v;
      }
      public int compareTo(edge e)
      {
          if(this.v>=e.v)
          return 1;
          else
          return -1;
  
      }
  }