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.BufferedOutputStream;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.HashMap;
import java.util.HashSet;
import java.util.LinkedList;
import java.util.StringTokenizer;


public class Pearls_in_a_Row {

	public static void main(String[] args) throws IOException {
		
		BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
		PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
		
		int n = Integer.parseInt(in.readLine());
		
		int[] a = new int[n];
		
		StringTokenizer st = new StringTokenizer(in.readLine());
		
		for (int i = 0; i < n; ++ i) {
			a[i] = Integer.parseInt(st.nextToken());			
		}
		
		int c = 0;
		int left = 0;		
		HashSet<Integer> S = new HashSet<>();
		LinkedList<Integer> l = new LinkedList<>();
		LinkedList<Integer> r = new LinkedList<>();
		for (int i = 0; i < n; ++ i) {			
			if (S.contains(a[i])) {
				S.clear();
				l.add(left+1);
				r.add(i+1);				
				left = i + 1;
				++ c;
			} else {
				S.add(a[i]);				
			}
		}
		
		if (c == 0) {
			out.println("-1");
		} else {
			out.println(c);
			int x=-1, y=1;
			while (l.size() > 1) {
				x = l.poll();
				y = r.poll();
				out.println(x + " " + y);
			}
			x = l.poll();
			y = a.length;
			out.println(x + " " + y);
		}
		
		out.close();
		
	}
	
}