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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class CF616D {

	public static void main(String[] args) throws IOException {
		BufferedReader bu = new BufferedReader(new InputStreamReader(System.in));
		
		String[] in = bu.readLine().split(" ");
		int elements = Integer.parseInt(in[0]);
		int k = Integer.parseInt(in[1]);
		
		in = bu.readLine().split(" ");
		
		int[] count = new int[(int) (Math.pow(10, 6)+1)];
		
		int[] array = new int[elements];
		int i = 0;
		for (String s : in) {
			array[i] = Integer.parseInt(s);
			i++;
		}
		
		int l = 0;
		int r = 0;
		
		int maxl = 0;
		int maxr = 0;
		int max = 0;
		int contains = 0;
		for (int j = 0; j < array.length; j++) {
			//System.out.println(l + " " + r);
			count[array[j]]++;
			if (count[array[j]]==1) {
				contains++;
			}
			
			while (contains>k) {
				count[array[l]]--;
				if (count[array[l]]==0)contains--;
				l++;
				
			}
			
			if (r-l>max) {
				maxl=l;
				maxr=r;
				max = r-l;
			}
			r++;
		}
			System.out.println((maxl+1) + " " + (maxr+1));

	}

}
