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.ArrayList;
import java.util.Scanner;

public class Dovet
{

    // ----------------------------------------------------------
    /**
     * Place a description of your method here.
     *
     * @param args
     */
    public static void main(String[] args)
    {
        // TODO Auto-generated method stub
        Scanner in = new Scanner(System.in);
        int pearlNum = in.nextInt();
        HashSet<Integer> segments = new HashSet<>();
        int count = 0;
        ArrayList<Integer> firsts = new ArrayList<>();
        ArrayList<Integer> lasts = new ArrayList<>();
        int first = 1;
        // int last = -1;
        for (int i = 0; i < pearlNum; i++)
        {
            int temp = in.nextInt();
            if (segments.contains(temp))
            {
                firsts.add(first);
                first = i + 2;
                lasts.add(i + 1);
                segments.clear();
                count++;
            }
            else
            {
                segments.add(temp);
            }

        }
        if (count == 0)
        {
            System.out.println("-1");
        }
        else
        {
            System.out.println(count);
            for (int i = 0; i < firsts.size() - 1; i++)
            {
                System.out.println(firsts.get(i) + " " + lasts.get(i));
            }
            System.out.println(firsts.get(firsts.size() - 1) + " " + pearlNum);
        }
    }

}
