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.util.*;
import java.io.*;
public class d {
	public static void main(String[] args) throws IOException{
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer(br.readLine());
		int n = Integer.parseInt(st.nextToken());
		int k = Integer.parseInt(st.nextToken());
		int[] nums = new int[n];
		st = new StringTokenizer(br.readLine());
		for(int i = 0;i<n;i++){
			nums[i] = Integer.parseInt(st.nextToken());
		}
		
		int p1 = 0;
		int p2 = 0;
		int count = 1;
		HashMap<Integer, Integer> counts = new HashMap<Integer, Integer>();
		counts.put(nums[0], 1);
		int max = 1;
		int pos = 1;
		while(p2 < n){
			if(count <= k){
				
				int len = p2-p1+1;
				if(len > max){
					max = len;
					pos = p1+1;
				}
				
				p2++;
				if(p2 == n){
					break;
				}
				if(counts.containsKey(nums[p2])){
					counts.put(nums[p2], counts.get(nums[p2])+1);
				}
				else{
					count++;
					counts.put(nums[p2], 1);
				}
			}
			else{
				int cc = counts.get(nums[p1]);
				if(cc == 1){
					count--;
					counts.remove(nums[p1]);
				}
				else{
					counts.put(nums[p1], counts.get(nums[p1])-1);
				}
				
				p1++;
				
			}
			
			if(count <= k){
				int len = p2-p1+1;
				if(len > max){
					max = len;
					pos = p1+1;
				}
			}
		}
		System.out.println(pos+" "+(pos+max-1));
	}
}
