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.util.*;

public class JavaApplication301 {

  
    public static void main(String[] args) {
       Scanner sc = new Scanner(System.in);
       int n = sc.nextInt();
       Set<Integer> set = new HashSet();
       ArrayList<Integer> list = new ArrayList();
       int index= 1;
       for(int i=0;i<n;i++){
           int a = sc.nextInt();
           if(set.contains(a)){
               set.clear();
               list.add(index);list.add(i+1);
               index=i+2;
           }else{
               set.add(a);
           }
       }
       if(list.isEmpty()) System.out.println(-1);
       else{
           list.remove(list.size()-1);
           list.add(n);
           System.out.println(list.size()/2);
       }
       for(int i=0;i<list.size()-1;i=i+2){
           System.out.println(list.get(i)+" "+list.get(i+1));
       }
       
    }
    
}
