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.InputStreamReader;
import java.io.PrintWriter;


public class D {
	public static void main(String[] args) throws IOException{
		BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
		PrintWriter out = new PrintWriter(System.out);
		String[] lin = in.readLine().split(" ");
		int len = Integer.parseInt(lin[0]);
		int k = Integer.parseInt(lin[1]);
		String[] line = in.readLine().split(" ");
		int[] ar = new int[len+1];
		for(int i = 1; i<=len; i++){
			ar[i] = Integer.parseInt(line[i-1]);
		}
		int p1 = 1;
		int p2 = 1;
		int bestLen = 0;
		int l = p1;
		int r = p2;
		int active = 1;
		int[] freq = new int[1000001];
		freq[ar[1]]=1;
		while(p1<=len && p2<=len){
	//	System.out.println(p1+" "+p2);
	//		System.out.println(p1+" "+p2+" "+active);
			if(active<=k){
				int curLen = p2-p1;
				if(curLen>bestLen){
					l = p1;
					r = p2;
					bestLen=curLen;
				}
				p2++;
				if(p2<=len){
				//	System.out.println(ar[p2]+" "+freq[ar[p2]]);
					if(freq[ar[p2]]==0){
						active++;
					}
					freq[ar[p2]]++;
				}
			}
			else{
				freq[ar[p1]]--;
				if(freq[ar[p1]]==0){
					active--;
				}
				p1++;
			}
		}
		System.out.println(l+" "+r);
		
	}
}