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 Haskell (GHC 7.10.3) (judge verdict: Wrong Answer)

-- Try AtCoder
-- author: Leonardone @ NEETSDKASU
 
import qualified Data.Map as Map
import Data.List (find)
import Data.Array (listArray, (!))

main = putStrLn . solve =<< getLine

solve xs = ans where
    a = listArray (1, length xs) xs
    h = Map.singleton (a ! 1) 1
    mn k = k - 1
    g i j m =
        let
            x = Map.adjust mn (a ! i) m
            mx = maximum $ Map.elems x
            l = j - i + 1
        in
            if l <= mx * 2 then
                (i + 1, j, x)
            else
                g (i + 1) j x
    f _ j _ | j > length xs = "-1 -1"
    f i j m =
        let
            x = Map.insertWith (+) (a ! j) 1 m
            mx = maximum $ Map.elems x
            l = j - i + 1
        in
            if l < mx * 2 then
                show i ++ " " ++ show j
            else
                let (ii, jj, xx) = g i j x
                in f ii (jj + 1) xx
    ans = f 1 2 h