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.io.*;
import java.util.*;
public final class longest_k_good
{
	static Scanner sc=new Scanner(System.in);
	static PrintWriter out=new PrintWriter(System.out);
	static int maxn=1000010;
	
	public static void main(String args[]) throws Exception
	{
		int n=sc.nextInt(),k=sc.nextInt();
		int[] a=new int[n],cnt=new int[maxn];
		for(int i=0;i<n;i++)
		{
			a[i]=sc.nextInt();
		}
		int c=0,p=0,maxl=0,maxr=0;
		for(int i=0;i<n;i++)
		{
			while(p<n)
			{
				int curr_elem=a[p];
				cnt[curr_elem]++;
				if(cnt[curr_elem]==1)
				{
					c++;
				}
				if(c>k)
				{
					cnt[a[p]]--;
					if(cnt[a[p]]==0)
					{
						c--;
					}
					break;
				}
				p++;
			}
			if(p-i>maxr-maxl && c<=k)
			{
				maxr=p;
				maxl=i;
			}
			cnt[a[i]]--;
			if(cnt[a[i]]==0)
			{
				c--;
			}
		}
		out.println(maxl+1+" "+maxr);
		out.close();
	}
}