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.io.*;
import java.util.*;
import java.lang.*;

public class Rextester{
    public static void main(String[] args)throws IOException{
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer(br.readLine());
        StringTokenizer st1 = new StringTokenizer(br.readLine());
        br.close();
        int n = Integer.parseInt(st.nextToken());
        int k = Integer.parseInt(st.nextToken());
        int[] array = new int[n];
        for(int i=0;i<n;i++){
            array[i]=Integer.parseInt(st1.nextToken());
        }
        int[] last = new int[1000005];
        int i = 0,j=0,max = 0,iMax = 0,jMax = 0;
        HashSet<Integer> set = new HashSet<Integer>();
        while(i<n&&j<n){
            set.add(array[i]);
            last[array[i]]=i;
            if(set.size()<=k){
                if(i-j>max){
                    max=i-j;
                    iMax=i;
                    jMax=j;
                }
            }
            else{
                while(set.size()>k){
                    if(last[array[j]]==j){
                        set.remove(array[j]);
                    }
                    j++;
                }
                if(i-j>max){
                    max=i-j;
                    iMax=i;
                    jMax=j;
                }
            }  
            i++;
        }
        System.out.println((jMax+1)+" "+(iMax+1));
    }
}