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: Accepted)

use std::io;
use std::io::prelude::*;

fn run(args: &Vec<String>) -> (i32, i32) {
    let s = args[0].to_string();
    for i in 0..s.len() {
        if s.len() >= 2 && i <= s.len() - 2 && s[i..i+1] == s[i+1..i+2] {
            return ((i+1) as i32, (i+2) as i32);
        } else if s.len() >= 3 && i <= s.len() - 3 && s[i..i+1] == s[i+2..i+3] {
            return ((i+1) as i32, (i+3) as i32);
        }
    }
    (-1, -1)
}

fn main() {
    let stdin = io::stdin();
    let args = stdin.lock().lines().map(|l| l.unwrap()).collect();
    let r = run(&args);
    println!("{} {}", r.0, r.1);
}