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 CF616D {
	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());
		st = new StringTokenizer(br.readLine());
		int[] aa = new int[n];
		for (int i = 0; i < n; i++)
			aa[i] = Integer.parseInt(st.nextToken());
		HashMap<Integer, Integer> map = new HashMap<>();
		int i = 0, j = 0, max = 0, l = 0, r = 0;
		while (j < n) {
			map.put(aa[j], map.getOrDefault(aa[j], 0) + 1);
			while (map.size() > k) {
				int cnt = map.get(aa[i]);
				if (cnt > 1)
					map.put(aa[i], cnt - 1);
				else
					map.remove(aa[i]);
				i++;
			}
			if (max < j - i + 1) {
				max = j - i + 1;
				l = i;
				r = j;
			}
			j++;
		}
		System.out.println((l + 1) + " " + (r + 1));
	}
}
