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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.HashSet;



public class Solution {

	public static void main(String args[]) throws NumberFormatException,
			IOException {

		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		int n = Integer.parseInt(br.readLine());

		String s[] = br.readLine().split(" ");

		int a[] = new int[n];

		for (int i = 0; i < n; i++)
			a[i] = Integer.parseInt(s[i]);

		int count = 0;

		Bag b = new Bag();
		Bag t = b;

		for (int i = 0; i < n; i++) {
			HashSet<Integer> h = new HashSet<Integer>();

			int j = i;
			for (; j < n; j++) {
				if (h.contains(a[j])) {
					Bag temp = b;
					b = new Bag();
					b.start = i;
					b.end = j;
					temp.next = b;
					count++;
					// System.out.println("ll " + a[j] + " " + j);
					break;
				} else
					h.add(a[j]);
			}
			i = j;
		}

		if (count != 0) {
			System.out.println(count);
			t = t.next;
			while (t.next != null) {
				System.out.println((t.start + 1) + " " + (t.end + 1));
				t = t.next;
			}

			System.out.println((t.start + 1) + " " + (n));
		} else
			System.out.println(-1);
	}
}

class Bag {

	int start;
	int end;
	Bag next;
}