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: Wrong Answer)

#[allow(unused_imports)] use proconio::{input, marker::{Bytes, Chars, Usize1, Isize1}};
#[allow(unused_imports)] use std::cmp::{min, max};
#[allow(unused_imports)] use superslice::Ext as _;

#[proconio::fastout]
fn main() {
	input! {
		cs: Chars,
	}

	let mut v = vec![[0usize; 26]; cs.len()];

	v[0][(cs[0] as u8 - 'a' as u8) as usize] = 1;
	for i in 1..cs.len() {
		v[i] = v[i-1];
		v[i][(cs[i] as u8 - 'a' as u8) as usize] += 1;
	}

	for end in 1..cs.len() {
		let mut nmax = 0;
		for i in 0..26 {
			nmax = max(nmax, v[end][i]);
		}
		if nmax > (end+1) / 2 {
			println!("{} {}", 0+1, end+1);
			return;
		}
	}
	for start in 1..cs.len()-2 {
		for end in start+1..cs.len() {
			let mut nmax = 0;
			for i in 0..26 {
				nmax = max(nmax, v[end][i]- v[start-1][i]);
			}
			if nmax > end / 2 {
				println!("{} {}", start+1, end+1);
				return;
			}
		}
	}
	println!("-1 -1");
}