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 (1.42.0) (judge verdict: Accepted)

use std::collections::HashMap;

use proconio::fastout;
use whiteread::parse_line;

#[fastout]
fn main() {
    let s: String = parse_line().unwrap();
    let s: Vec<char> = s.chars().collect();

    let mut map: HashMap<char, Vec<usize>> = HashMap::new();
    for i in 0..s.len() {
        let c = s[i];
        let entry = map.entry(c).or_insert(vec![]);
        entry.push(i);
    }

    for c in (b'a'..=b'z').map(char::from) {
        if !map.contains_key(&c) || map[&c].len() == 1 {
            continue;
        }

        let mut left: usize = 0;
        let mut right: usize = map[&c].len() - 1;
        let mut removed = 0;

        loop {
            if left == right {
                break;
            }

            if (map[&c].len() - removed) * 2 > (map[&c][right] - map[&c][left] + 1) {
                println!("{} {}", map[&c][left] + 1, map[&c][right] + 1);
                return;
            }

            if map[&c][left + 1] - map[&c][left] > map[&c][right] - map[&c][right - 1] {
                left += 1;
            } else {
                right -= 1
            }
            removed += 1;
        }
    }

    println!("-1 -1");
}