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
//package bnnb;

import java.util.Scanner;

public class moode {

	public static void main(String[] args) {
		Scanner s=new Scanner(System.in);
		   int n,k;
		    n=s.nextInt();
		    k=s.nextInt();
		    int a[]=new int[n];int c=0,p=0,l=0,r=-55,c2=0,mx=0;
		    int a1[]=new int[1000001];
		    for(int i=0;i<n;i++)a[i]=s.nextInt();
		
		    for(int i=0;i<n;){
		        
		       
		         if(a1[a[i]]==0&&c==k){
		               if(c2>mx){
		                mx=c2;
		                l=p;
		                r=i-1;
		                
		            }
		                a1[a[p]]--;
		                if(a1[a[p]]==0){c--;}
		                p++;
		          c2--;
		         }else  {
		              c2++;
		        a1[a[i]]++;
		        if(a1[a[i]]==1)
		       c++;
		        i++;
		        }
		        
				    

		        
		    }//System.out.println(p);
		    if(c2>mx) {r=n-1;if(l+1!=0)l=p;}
		    System.out.println((l+1)+" "+(r+1));
}

}