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.util.HashSet;
import java.util.LinkedList;
import java.util.StringTokenizer;

public class C {
  public static void main(String[] args) throws Exception {
    BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
    int n = Integer.parseInt(in.readLine());
    StringTokenizer str = new StringTokenizer(in.readLine());
    LinkedList<Integer> list = new LinkedList<>();
    HashSet<Integer> set = new HashSet<>();
    list.add(0);
    for (int i = 0; i < n; i++) {
      int k = Integer.parseInt(str.nextToken());
      if (set.contains(k)) {
        list.add(i + 1);
        set.clear();
      } else {
        set.add(k);
      }
    }
    if (list.size() == 1) {
      System.out.println(-1);
    } else {
      System.out.println(list.size() - 1);
      if (list.peekLast() != n) {
        list.pollLast();
        list.addLast(n);
      }
      for (int i = list.pollFirst(); !list.isEmpty(); i = list.pollFirst())
        System.out.println((i + 1) + " " + list.peekFirst());
    }
  }
}