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.util.*;
import java.io.*;

public class CF616D {
	static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
	static BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
	static int[] arr;
	static int n, k;

	public static void main(String[] args) throws IOException {
		String[] first = br.readLine().split(" ");
		n = Integer.parseInt(first[0]);
		k = Integer.parseInt(first[1]);
		TreeMap<Integer, Integer> tree = new TreeMap<>();
		// TreeSet<Integer> tree = new TreeSet<>();

		first = br.readLine().split(" ");
		arr = new int[n];
		for(int i = 0; i < first.length; i++) {
			arr[i] = Integer.parseInt(first[i]);
		}

		int left = 0;
		int right = 1;

		int leftPointer = 0;
		int rightPointer = 0;

		while(tree.size() < k && rightPointer < n) {
			if(!tree.containsKey(arr[rightPointer])) {
				tree.put(arr[rightPointer++], 1);
			}
			else {
				int count = tree.get(arr[rightPointer]) + 1;
				tree.replace(arr[rightPointer++], count);
			}

			right = rightPointer;
		}

		while(rightPointer < n && leftPointer < n) {
			if(tree.containsKey(arr[rightPointer])) {
				int count = tree.get(arr[rightPointer]) + 1;
				tree.replace(arr[rightPointer++], count);
			}
			else {
				tree.put(arr[rightPointer++], 1);
				while(true) {
					int count = tree.get(arr[leftPointer]) - 1;
					if(count == 0) {
						tree.remove(arr[leftPointer++]);
						break;
					}
					else {
						tree.put(arr[leftPointer++], count);
					}
				}
			}

			if(right - left <= rightPointer - leftPointer) {
				left = leftPointer;
				right = rightPointer;
			}
		}

		System.out.println((left+1) + " " + (right));

	}

	// public static int recursion(int L, int R, TreeSet<Integer> t) {
	// 	if(t.size() <= k || R >= n || L >= n) {
	// 		return R - L;
	// 	}

	// 	TreeSet<Integer> tree = (TreeSet<Integer>)t.clone();

	// 	boolean containsLeft = tree.contains(arr[L]);
	// 	if(containsLeft) {
	// 		tree.remove(arr[L]);
	// 	}
	// 	int leftTree = recursion(L+1, R, tree);
	// 	if(containsLeft) {
	// 		tree.add(arr[L]);
	// 	}

	// 	boolean containsRight = tree.contains(arr[L]);
	// 	if(containsRight) {
	// 		tree.remove(arr[R]);
	// 	}
	// 	int rightTree = recursion(L+1, R, tree);
	// 	if(containsRight) {
	// 		tree.add(arr[R]);
	// 	}

	// 	return Math.max(leftTree, rightTree);
	// }
}