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.HashSet;
import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        HashSet<Integer> s = new HashSet<>();
        int ans = 0;
        int head[] = new int[300005];
        int tail[] = new int[300005];
        for (int j = 0, i = 0; j < n; j++) {
            Integer a = in.nextInt();
            if (s.contains(a)) {
                s.clear();
                head[ans] = i + 1;
                tail[ans++] = j + 1;
                i = j + 1;
            } else {
                s.add(a);
            }
        }
        System.out.println(ans == 0 ? -1 : ans);
        for (int i = 0; i < ans; i++) {
            System.out.println(head[i] + " " + (i == ans-1 ? n : tail[i]));
        }
    }
}

				      	  		     	 	  		 	 	