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 sol{
    public static void main(String args[]){
        Scanner sc=new Scanner(System.in);
        int n=sc.nextInt();
        HashSet<Integer> hs=new HashSet<>();
        int arr[]=new int[n];
        int cntr=0;
        for(int i=0;i<n;i++){
            arr[i]=sc.nextInt();
        }
        ArrayList<Integer> al=new ArrayList<>();
        int pi=1;
        for(int i=0;i<n;i++){
            if(hs.contains(arr[i])){
                cntr++;
                al.add(pi);
                al.add(i+1);
                pi=i+2;
                hs.clear();
            }
            else{
                hs.add(arr[i]);
            }
            //System.out.println(hs);
        }
        if(cntr==0)System.out.println(-1);
        else{
        al.remove(al.size()-1);
        al.add(n);
        System.out.println(cntr);
        for(int i=0;i<al.size();i+=2)
            System.out.println(al.get(i)+" "+al.get(i+1));}
    }
}