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

public class SubarraySum2 {

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

		BufferedReader in = new BufferedReader(new InputStreamReader(System.in));

		StringTokenizer st = new StringTokenizer(in.readLine());
		int n = Integer.parseInt(st.nextToken());
		int k = Integer.parseInt(st.nextToken());
		
		int[] array = new int[n];
		int max = 0;
		st = new StringTokenizer(in.readLine());
		for(int i = 0; i < n; i ++) {
			array[i] = Integer.parseInt(st.nextToken());
			max = Math.max(max, array[i]);
		}
		
		
		int[] freq = new int[max+1];
		int i = 0, j = 0;
		int length = 0;
		
		int left = 1, right = 1;
		
		while(j < n) {
			while(j < n && length <= k) {
				if(freq[array[j]] ++ == 0) {
					length ++;
				}
				j ++;
			}
			
			// when break, the answer is one more than the actual ans (only if j is not at the end)
			int newI = i+1;
			int newJ = j == n && length <= k ? j: j-1;
			
			if(newJ - newI > right - left) {
				left = newI;
				right = newJ;
			}
			
			// remove while there is more elements than length
			while(length > k) {
				if(--freq[array[i]] == 0) {
					length --;
				}
				i ++;
			}
		}
		
		System.out.println(left + " " + right);
				
		in.close();
	}
}
