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 longestkgoodsegment {
	public static void main(String[] args) throws IOException {
		BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer(f.readLine());
		int n = Integer.parseInt(st.nextToken());
		int k = Integer.parseInt(st.nextToken());
		int[] arr = new int[n];
		st = new StringTokenizer(f.readLine());
		for(int i = 0; i < n; i ++) arr[i] = Integer.parseInt(st.nextToken());
		int longest = 0;
		int left = 0;
		int right = 0;
		int tempLeft = 0;
		int[] count = new int[1000001];
		int kTemp = 0;
		for(int i = 0; i < n; i ++) {
			count[arr[i]] ++;
			if(count[arr[i]] == 1) kTemp ++;
			while(kTemp > k) {
				count[arr[tempLeft]] --;
				if(count[arr[tempLeft]] == 0) kTemp --;
				tempLeft ++;
			}
			if(i - tempLeft + 1 > longest) {
				longest = i - tempLeft + 1;
				left = tempLeft;
				right = i;
			}
		}
		System.out.println((left + 1) + " " + (right + 1));
	}
}
