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
//package problem2;

import java.util.Scanner;

public class Main {

	private static final int MAX_VALUE = 1000000;
	
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		// Read the input.
		int n = sc.nextInt();
		int k = sc.nextInt();
		// Declare and read the array.
		int[] array = new int[n];
		for (int i = 0; i < n; ++i) {
			array[i] = sc.nextInt();
		}
		
		// Iterate with two pointers.
		int freq[] = new int[MAX_VALUE];
		int p = -1, distinct = 0;
		// Solution values.
		int optimalLen = 0, optimalLeft = 0, optimalRight = 0;
		for (int i = 0; i < n; ++i) {
			// We have to remove the previously element.
			if (i > 0) {
				--freq[array[i - 1]];
				if (freq[array[i - 1]] == 0) {
					--distinct;
				}
			}
			// Try to expand.
			while (p + 1 < n) {
				// If we add the next element, compute the number of
				// distinct values.
				int nextDistinct = distinct;
				if (freq[array[p + 1]] == 0) {
					++nextDistinct;
				}
				// Check if we expand.
				if (nextDistinct > k) {
					break;
				}
				// Expand.
				++p;
				++freq[array[p]];
				distinct = nextDistinct;
			}
			// Try to update the current solution.
			if (p - i + 1 > optimalLen) {
				optimalLen = p - i + 1;
				optimalLeft = i + 1;
				optimalRight = p + 1;
			}
		}
		
		System.out.println(optimalLeft + " " + optimalRight);
	}
}
