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.*;

public class Edu63 {
    Scanner in = new Scanner(System.in);

    public static void main(String[] args) {
        new Edu63().run();
    }

    private void run() {
        int count = 0;
        int r = 1;
        Set<Long> buf = new HashSet<>(100);
        List<Integer> a = new ArrayList<>();
        List<Integer> b = new ArrayList<>();

        int n = in.nextInt();
        for (int i = 1; i <= n; i++) {
            long number = in.nextInt();
            if (buf.contains(number)) {
                count++;
                a.add(r);
                b.add(i);
                r = i + 1;
                buf.clear();
            } else {
                buf.add(number);
            }
        }
        int size = b.size();
        if (size > 0) b.set(size - 1, n);

        System.out.println(count == 0 ? -1 : count);
        for (int i = 0; i < a.size(); i++) {
            System.out.println(a.get(i) + " " + b.get(i));
        }
    }
}