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 Rust (1.15.1) (judge verdict: Wrong Answer)

use std::collections::HashMap;

fn main() {
    let s: String = {
        let mut buffer = String::new();
        std::io::stdin().read_line(&mut buffer).unwrap();

        buffer.trim().to_string()
    };

    let mut hashmap: HashMap<char, Vec<usize>> = HashMap::new();

    for (i, c) in s.chars().enumerate() {
        if hashmap.contains_key(&c) {
            let v = hashmap.get_mut(&c).unwrap();
            v.push(i);
        } else {
            let v = vec![i];
            hashmap.insert(c, v);
        }
    }

    let mut unbalance = None;

    'outer: for (_, vec) in hashmap.clone() {
        for i in 0..vec.len() {
            for j in i..vec.len() {
                if i != j && ((vec[j] - vec[i]) / 2) <= (j - i) {
                    if i != 0 || j != (vec.len() - 1) {
                        unbalance = Some((vec[i], vec[j]));
                        break 'outer;
                    }
                }
            }
        }
    }

    if let Some((i, j)) = unbalance {
        println!("{} {}", i + 1, j + 1);
    } else {
        println!("-1 -1");
    }
}