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 Python (2.7.6) (judge verdict: Wrong Answer)

import sys
from collections import deque
import copy
import math
def get_read_func(fileobject):
    if fileobject == None :
        return raw_input
    else:
        return fileobject.readline


def main():
    if len(sys.argv) > 1:
        f = open(sys.argv[1])
    else:
        f = None
    read_func = get_read_func(f);
    input_raw = read_func().strip().split()
    S = input_raw[0]


    for k in range(2, 4):
        for i in range(len(S) - k):
            dic = {}
            for j in range(k):
                if S[i + j] not in dic:
                    dic[S[i + j]] = 1
                else:
                    dic[S[i + j]] += 1
            for j in range(k):
                if dic[S[i + j]] >= k/2 + 1:
                    print i + 1, k + i + 1
                    return
    print -1, -1

##
##    dic = {}
##    for i in range(len(S)):
##        if S[i] not in dic:
##            dic[S[i]] = [i]
##        else:
##            dic[S[i]].append(i)
##    for s in dic:
##        for i in range(len(dic[s])):

if __name__ == '__main__':
    main()