Problem: 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

Solution (in JAVA):

Solution:
import java.util.Arrays;
import java.util.HashSet;
import java.util.Scanner;
import java.util.Set;

public class Educational6B {
	public static void main(String[] args){
		Scanner sc = new Scanner(System.in);
		Set s=new HashSet();
		int n=sc.nextInt();
		int []num=new int[n];
		int ans=0;
		int len=s.size();
		int [][]lr=new int[300001][2];
		for(int i=0;i<300001;i++)
			Arrays.fill(lr[i], 0);
		lr[ans][0]=0;
		for(int i=0;i<n;i++){
			num[i]=sc.nextInt();
			s.add(num[i]);
			if(s.size()>len){
				len=s.size();
			}
			else{
				s.clear();
				len=s.size();
				lr[ans][1]=i;
				ans++;
				lr[ans][0]=i+1;
			}
		}
		if(lr[ans][1]==0&&ans!=0)
			lr[ans-1][1]=n-1;
		if(ans==0)
			System.out.println(-1);
		else{
			System.out.println(ans);
			for(int i=0;i<ans;i++){
				System.out.println((lr[i][0]+1)+" "+(lr[i][1]+1));
			}
		}
	}
}