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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.Comparator;
import java.util.StringTokenizer;
import javafx.util.Pair;

public class Solve3 {

    public static void main(String[] args) throws IOException {
        PrintWriter pw = new PrintWriter(System.out);
        new Solve3().solve(pw);
        pw.flush();
        pw.close();
    }

    public void solve(PrintWriter pw) throws IOException {
        FastReader sc = new FastReader();
        int n = sc.nextInt(), k = sc.nextInt();
        int[] a = new int[n + 1];
        for (int i = 1; i <= n; i++) {
            a[i] = sc.nextInt();
        }
        final int MAX = (int) 1E+6;
        int[] freq = new int[MAX + 1];
        int l = 1, r = 1, max = 0, lans = 1, rans = 1;
        int j = 0;
        while (r <= n) {
            if (freq[a[r]] == 0) {
                j++;
                while (j > k) {
                    --freq[a[l]];
                    if (freq[a[l]] == 0) {
                        --j;
                    }
                    ++l;
                }
            }
            if (r - l + 1 > max) {
                max = r - l + 1;
                lans = l;
                rans = r;
            }
            ++freq[a[r]];
            ++r;
        }
        pw.println(lans + " " + rans);
    }

    static public class FastReader {

        BufferedReader br;
        StringTokenizer st;

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

        public String next() throws IOException {
            if (st == null || !st.hasMoreTokens()) {
                st = new StringTokenizer(br.readLine());
            }
            return st.nextToken();
        }

        public int nextInt() throws IOException {
            return Integer.parseInt(next());
        }

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

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

        public String nextLine() throws IOException {
            return br.readLine();
        }
    }
}
