Problem:
Score : 400 points

Problem Statement

Given a string t, we will call it unbalanced if and only if the length of t is at least 2, and more than half of the letters in t are the same. For example, both voodoo and melee are unbalanced, while neither noon nor a is.

You are given a string s consisting of lowercase letters. Determine if there exists a (contiguous) substring of s that is unbalanced. If the answer is positive, show a position where such a substring occurs in s.

Constraints

2 ≦ |s| ≦ 10^5

s consists of lowercase letters.

Partial Score

200 points will be awarded for passing the test set satisfying 2 ≦ N ≦ 100.

Input

The input is given from Standard Input in the following format:

s

Output

If there exists no unbalanced substring of s, print -1 -1.

If there exists an unbalanced substring of s, let one such substring be s_a s_{a+1} ... s_{b} (1 ≦ a < b ≦ |s|), and print a b. If there exists more than one such substring, any of them will be accepted.

Sample Input 1

needed

Sample Output 1

2 5

The string s_2 s_3 s_4 s_5 = eede is unbalanced. There are also other unbalanced substrings. For example, the output 2 6 will also be accepted.

Sample Input 2

atcoder

Sample Output 2

-1 -1

The string atcoder contains no unbalanced substring.

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.*;

public class Main
{
  public static void main(String[] args)
  {
    Scanner sc = new Scanner(System.in);
    String str = sc.nextLine();
    List<String> list = Arrays.asList(str.split(""));
    sc.close();
    
    if(list.size() <= 1)
    {
      System.out.println("-1 -1");
      return;
    }
    
    Set<String> set = new HashSet<String>(list);
    Map<String,Long> map = new HashMap<String,Long>(set.size());
    
    long max = 0;
    String max_s = "";
    long max2 = -1;
    for(String s : set)
    {
      map.put(s,new Long(0));
    }
    for(String s:list)
    {
      long val = map.get(s) + 1;
      map.put(s,val);
      if(val > max)
      {
        max2 = max;
        max_s = s;
        max = val;
      }
      else if(val == max)
      {
        max2 = val;
      }
    }
    if(max2 != max && max >= (str.length() - max))
      System.out.println((str.indexOf(max_s) + 1) + " " + (str.lastIndexOf(max_s) + 1));
    else
      System.out.println("-1 -1");
    
    
  }
}