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 Solution{
			

	static class Pair{
		int x,y;
		Pair(int x,int y)
		{
			this.x=x;
			this.y=y;
		}
	}

	public static void main(String args[]) throws IOException
	{
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		PrintWriter out=new PrintWriter(System.out);

		int n=Integer.parseInt(br.readLine());
		int a[]=new int[n];

		String s[]=br.readLine().split(" ");
		for(int i=0;i<n;i++)
			a[i]=Integer.parseInt(s[i]);

		Set<Integer> set=new HashSet<Integer>();

		ArrayList<Pair> ans=new ArrayList<Pair>();

		int i=0;
		int left=1;
		while(i<n)
		{
			if(!set.contains(a[i]))
				set.add(a[i]);
			else
			{
				set.clear();
				Pair x=new Pair(left,i+1);
				ans.add(x);
				left=i+2;
			}
			i++;
		}		

		int len=ans.size();
		if(len!=0)
		{
			out.println(len);

			for(i=0;i<len;i++)
			{
				Pair x=ans.get(i);

				if(i==len-1)
					out.println(x.x+" "+n);
				else
					out.println(x.x+" "+x.y);
			}
		}
		else
			out.println(-1);



		out.flush();
		out.close();
			

	}
}