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:
//package CodeForces.EducationalRound5;

import java.io.BufferedOutputStream;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;

/**
 * Created by ilya on 13.01.16.
 */

public class D616 {
    public static void main(String[] args) {
        sc sc = new sc();
        PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
        int n = sc.nextInt();
        int k = sc.nextInt();
        int[] m = new int[n];
        for (int i = 0; i < n; i++) {
            m[i] = sc.nextInt();
        }
        int[] count = new int[1000001];
        int r = 0, l = 0, in = 0, j = 0;
        for (int i = 0; i < n; i++) {
            if (i > 0) {
                count[m[i - 1]]--;
                if (count[m[i - 1]] == 0) in--;
            }
            while (j < n && in <= k) {
                count[m[j]]++;
                if (count[m[j]] == 1) in++;
                if (in > k) {
                    count[m[j]]--;
                    in--;
                    break;
                } else j++;
            }

            if (j - i > r - l) {
                r = j;
                l = i;
            }
        }
        out.println((l + 1) + " " + (r));
        out.close();
    }

    public static class sc {
        BufferedReader br;
        StringTokenizer st;

        public sc() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        Integer nextInt() {
            return Integer.parseInt(next());
        }

        long nextLong() {
            return Long.parseLong(next());
        }

        double nextDouble() {
            return Double.parseDouble(next());
        }

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }

}