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.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;

public class LongestKGoodSegment {
    static int begin = 1, end = 1, max = -1;
    static int[] diff;
    static int[] a;
    static int[] temp;
    public static void main(String[] args) throws Exception {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter pw = new PrintWriter(System.out);
        String[] s;
        s = br.readLine().split(" ");
        int n = Integer.parseInt(s[0]), k = Integer.parseInt(s[1]), countDiff = 0, i = 0, j = 0, l = 0;
        s = br.readLine().split(" ");
        a = new int[n];
        for (i = 0; i < n; i++) {
           a[i] = Integer.parseInt(s[i]);
           if (max < a[i])
               max = a[i];
        }
        i = 0;
        diff = new int[max + 1];
        max = -1;
        for (; j < n; j++) {
            diff[a[j]]++;
            if (diff[a[j]] == 1) countDiff++;
            if (countDiff > k) {
                while (countDiff > k) {
                    diff[a[i]]--;
                    if (diff[a[i]] == 0) countDiff--;
                    i++;
                }
            }
            if (countDiff <= k)
                setValue(i, j);
            
        }
        pw.println(begin + " " + end);
        pw.close();
    }
    
    static void setValue (int i, int j) {
        int count = j - i + 1;
        if (max < count) {
            max = count;
            begin = i + 1;
            end = j + 1;
        }
    }
}