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



import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;

    public class Training {

        static Integer a[]= new Integer[1000001]; 
            public static void main(String[] args) throws IOException {
 		Scanner input = new Scanner(System.in);
               HashSet<Integer> a= new HashSet<Integer>();
                int  b[]= new int[300001];
                int n = input.nextInt();
                int c = 0 ; 
                int  p =1 ;
                for(int i = 0;i<n;i++){
                    int x = input.nextInt();
                    if(a.contains(x)){
                        a.clear();
                        c++;
                        b[p] = i+1 ; 
                        p =i+2;
   
                    }else{
                        a.add(x);
                    }
                } 
                
                if(c==0){
                    System.out.println(-1);
                }else{
                    System.out.println(c);
                  for(int i = 300000;i>=1;i-- ){
                        if(b[i]!=0){
                            b[i]= n;
                            break;
                        }
                    }
                    for(int i =1;i<=300000;i++){
                        if(b[i]!=0){
                            System.out.println(i+" "+b[i]);
                        }
                    }
                }
                
               
                
        } // end Main



    } // end Class
