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.InputStream;
import java.io.InputStreamReader;
import java.util.Deque;
import java.util.HashMap;
import java.util.LinkedList;
import java.util.Scanner;


public class Main {

	/**
	 * @param args
	 * @throws IOException 
	 */

	public static void main(String[] args) throws IOException {
		// TODO Auto-generated method stub
	//	Scanner scan = new Scanner(System.in);
		Deque<Integer> dq = new LinkedList<Integer>();
		 BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
         String s[] = br.readLine().split(" ");
		HashMap<Integer,Integer> occ = new HashMap<Integer,Integer>();
		int n = Integer.parseInt(s[0]),k = Integer.parseInt(s[1]);
		int a[] = new int[n];
		String s1[] = br.readLine().split(" ");
		for(int i = 0; i < n; i++)
			a[i] = Integer.parseInt(s1[i]);
		int l = 0,r = 0,sz = 0,max = 0;
		for(int i = 0; i < n; i++) {
			if(occ.containsKey(a[i])) {
				dq.addLast(a[i]);
				int x = occ.get(a[i]);
				occ.put(a[i], x + 1);
				sz++;
			}
			else {
				if(sz < k) {
					dq.addLast(a[i]);
					occ.put(a[i], 1);
					sz++;
				}
				else {
					while(occ.size() >= k) {
						int e = dq.peekFirst();
						dq.pollFirst();
						int t = occ.get(e);
						if(t == 1)
							occ.remove(e);
						else
							occ.put(e, t - 1);
					}
					
					dq.addLast(a[i]);
					if(occ.containsKey(a[i])) {
						int x = occ.get(a[i]);
						occ.put(a[i], x + 1);
					}
					else {
						occ.put(a[i], 1);
					}
					sz = dq.size();
					
				}
			}
			if(sz > max) {
				max = sz;
				l = i - sz + 2;
				r = i + 1;
			}
			
		}
		System.out.println(l + " " + r);
	

	}

}