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.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;

public class D {

	
	public static void main(String[] args) throws IOException
	{	
		Scanner sc = new Scanner(System.in);
		PrintWriter out = new PrintWriter(System.out); 
		int n = sc.nextInt(), k = sc.nextInt();
		int[] a = new int[n]; for(int i = 0; i < n; i++) a[i] = sc.nextInt();
		int[] occ = new int[1000001];
		int c = 0, i = 0, j = 0, max = 0, s = 0, e = 0;
		while(i < n && j < n)
		{
//			System.out.println(i + " " + c);
			occ[a[i]]++;
			if(occ[a[i]] > 1)
			{
				if(i - j + 1 > max)
				{
					max = i - j + 1;
					s = j; e = i;
				}
				i++; continue;
			}
			if(c < k)
			{
				if(i - j + 1 > max)
				{
					max = i - j + 1;
					s = j; e = i;
				}
				c++; i++; continue;
			}
			if(i - j > max)
			{
				max = i - j;
				s = j; e = i-1;
			}
			while(j < n)
			{
				occ[a[j]]--;
				if(occ[a[j]] == 0) break;
				j++;
			}
			j++; i++;
		}
		out.println((s + 1) + " " + (e + 1));
		out.flush();
		out.close();
	}
	
	static class Scanner 
	{
		StringTokenizer st;
		BufferedReader br;

		public Scanner(InputStream System){	br = new BufferedReader(new InputStreamReader(System));	}
		public String next() throws IOException 
		{
			while (st == null || !st.hasMoreTokens()) 
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}
		public String nextLine()throws IOException{return br.readLine();}
		public int nextInt() throws IOException {return Integer.parseInt(next());}
		public double nextDouble() throws IOException {return Double.parseDouble(next());}
		public char nextChar()throws IOException{return next().charAt(0);}
		public Long nextLong()throws IOException{return Long.parseLong(next());}
		public boolean ready() throws IOException{return br.ready();}
		public void waitForInput(){for(long i = 0; i < 3e9; i++);}
	}
}