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 static java.lang.Integer.MIN_VALUE;
import static java.lang.Integer.parseInt;

public class D_Longest_k_Good_Segment__Sol2 {
    static int n,k,ar[],count;
    static int freq[];
    public static void main(String[] args) throws IOException {
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
        StringTokenizer tk = new StringTokenizer(in.readLine());
        n = parseInt(tk.nextToken());
        k = parseInt(tk.nextToken());
        ar = new int[n];
        freq = new int[1000010];
        tk = new StringTokenizer(in.readLine());
        for (int i = 0; i < n; i++) {
            ar[i] = parseInt(tk.nextToken());
        }

        int p = 0;
        int l = 0,r = 0;
        int ans = MIN_VALUE;
        for (int i = 0; i < n; i++) {
            freq[ar[i]]++;
            if(freq[ar[i]] == 1)
                count++;
            if(count <= k){
                if(i - p + 1 > ans){
                    ans = i - p + 1;
                    l = p;
                    r = i;
                }
            }
            else {
                while (p < i && count > k){
                    freq[ar[p]]--;
                    if(freq[ar[p]] == 0)
                        count--;
                    p++;
                }
            }
        }
        ++l; ++r;
        out.println(l+" "+r);
        out.flush();
    }

}