The array a with n integers is given. Let's call the sequence of one or more consecutive elements in a segment. Also let's call the segment k-good if it contains no more than k different values.

Find any longest k-good segment.

As the input/output can reach huge size it is recommended to use fast input/output methods: for example, prefer to use scanf/printf instead of cin/cout in C++, prefer to use BufferedReader/PrintWriter instead of Scanner/System.out in Java.

Input

The first line contains two integers n, k (1 ≤ k ≤ n ≤ 5·105) — the number of elements in a and the parameter k.

The second line contains n integers ai (0 ≤ ai ≤ 106) — the elements of the array a.

Output

Print two integers l, r (1 ≤ l ≤ r ≤ n) — the index of the left and the index of the right ends of some k-good longest segment. If there are several longest segments you can print any of them. The elements in a are numbered from 1 to n from left to right.

Examples

Input

5 5
1 2 3 4 5


Output

1 5


Input

9 3
6 5 1 2 3 2 1 4 5


Output

3 7


Input

3 1
1 2 3


Output

1 1
/*
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 * To change this template file, choose Tools | Templates
 * and open the template in the editor.
 */
import java.util.ArrayList;
import java.util.Collection;
import java.util.Collections;
import java.util.Scanner;
import javafx.util.Pair;

/**
 *
 * @author ahmadla
 */
public class KEFA1 {

    /**
     * @param args the command line arguments
     */
    public static void main(String[] args) {
        // TODO code application logic here
        Scanner in =new Scanner(System.in);
        long n=in.nextLong();
        long t=in.nextLong();
        ArrayList arr=new ArrayList();
        
        for (int i = 0; i < n; i++) {
       
            arr.add( in.nextInt());
        }
       
   int a[]=new int[1000000];
 
 int r=0,l=0;

  boolean f1=false;
 long ans=0,sum=0;
 int ansr=0;
 int ansl=0;
   int f=0;
   int max=0;
     while(r<n)
     {  if(a[(Integer)arr.get(r)]==0)
         f++;
 a[(Integer)arr.get(r)]++;
        
     if(f<=t)
     {
        
    if(r-l>max)
    {ansr=r;
    ansl=l;
    max=r-l;
   
    }
   
     }
      
     else
     { 
      a[  (Integer)arr.get(l)]--;
   if(a[  (Integer)arr.get(l)]==0)
      f--;
     l++;
   
     }
       r++;
     }
        System.out.println((ansl+1)+" "+(ansr+1));
    }
    
}
