Problem: 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

Solution (in JAVA):

Solution:
import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;

public class main {
    public static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    public static PrintWriter pw = new PrintWriter(System.out);
    public static String line;
    public static StringTokenizer st;

    public static void main(String[] args) throws Exception{
        int N = Integer.parseInt(br.readLine());
        st = new StringTokenizer(br.readLine());
        int[] A = new int[N];
        for(int i = 0; i < N; i++){
            A[i] = Integer.parseInt(st.nextToken());
        }
        TreeMap<Integer, Integer> d = new TreeMap<Integer, Integer>();
        ArrayList<Pair> out = new ArrayList<Pair>();

        int last = 0;
        for(int i = 0; i < N; i++){
            if(d.containsKey(A[i])){
                out.add(new Pair(last+1, i+1));
                last = i+1;
                d = new TreeMap<Integer, Integer>();
            } else{
                d.put(A[i], i);
            }
        }
        if(out.size() == 0){
            pw.print("-1\n");
        } else{
            Pair l = out.get(out.size()-1);
            l.b = N;
            pw.print(out.size() + "\n");
            for(Pair P : out){
                pw.print(P.a + " " + P.b + "\n");
            }
        }
        pw.close();
        br.close();
    }
}

class Pair implements Comparable<Pair>{
    public int a;
    public int b;

    Pair(int a, int b){
        this.a = a;
        this.b = b;
    }

    public int compareTo(Pair P){
        if(this.a < P.a){
            return -1;
        } else if(this.a > P.a){
            return 1;
        } else if(this.b < P.b){
            return -1;
        } else if(this.b > P.b){
            return 1;
        } else{
            return 1;
        }
    }
}