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.*;
public class Solution
{
    public static void main(String ag[])throws Exception
    {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        int i,j,k;
        String str[]=br.readLine().split(" ");
        int N=Integer.parseInt(str[0]);
        int K=Integer.parseInt(str[1]);
        
        String arr[]=br.readLine().split(" ");
        int A[]=new int[N];
        for(i=0;i<N;i++)
        {
            A[i]=Integer.parseInt(arr[i]);
        }
        
        int freq[]=new int[(int)1e6+1];
        freq[A[0]]++;
        int ans=1;
        int max=1;
        int L=1;
        int R=1;
        i=0;
        j=0;
        
        while(j<N)
        {
            if(j<i)
            {
                j++;
                continue;
            }
            if(ans<=K)
            {
                if(j==N-1)
                break;
                
                j++;
                freq[A[j]]++;
                if(freq[A[j]]==1)
                ans++;
            }
            else 
            {
                i++;
                freq[A[i-1]]--;
                if(freq[A[i-1]]==0)
                ans--;
            }
            if(j-i+1>max&&ans<=K)
            {
                max=j-i+1;
                L=i+1;
                R=j+1;
            }
        }
        System.out.println(L+" "+R);
    }
}