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.*;
import java.io.*;
public class cf{
    public static void main(String args[])throws Exception{
     Scanner sc=new Scanner(new BufferedReader(new InputStreamReader(System.in)));
	int n=sc.nextInt();
	ArrayList<String> ans=new ArrayList<>();
	HashMap<Long,Integer> indices=new HashMap<>();
	for(int i=0;i<n;i++){
	    long x=sc.nextLong();
	    if(!indices.containsKey(x))
		indices.put(x,i+1);
	    else{
		ans.add(""+indices.get(x)+" "+(i+1));
		indices=new HashMap<>();
	    }
	}
	if(ans.size()==0)System.out.println("-1");
	else{
	    int curr=1;
	    System.out.println(ans.size());
	    String y=ans.remove(ans.size()-1);
	    ans.add(y.substring(0,y.indexOf(' '))+" "+n);
	    for(String x:ans){
		System.out.println(""+curr+x.substring(x.indexOf(' ')));
		curr=Integer.parseInt(x.substring(x.indexOf(' ')+1))+1;
	    }
	}
    }
}
