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 PyPy3 (2.4.0) (judge verdict: Wrong Answer)

s = list(map(lambda x: ord(x) - ord('a'), list(input())))
def solve(s):
    ans = ["-1", "-1"]
    for c in range(26):
        d = []
        for idx, v in enumerate(s):
            if v == c:
                d.append(idx)
        if len(d) >= 2:
            for right in range(1, len(d)):
                for left in range(right-1):
                    whole = d[right] - d[left] + 1
                    cnt = right - left + 1
                    if cnt > whole // 2:
                        if cnt == 2 and whole == 2:
                            if 0 < left:
                                left -= 1
                            elif right < len(s) - 1:
                                right += 1
                            else:
                                return ["-1", "-1"], None
                        else:
                            ans = [str(left+1), str(right+1)]
                            return ans, s[left:right+1]
    return ans, None
ans, l = solve(s)
# print(l)
print(" ".join(ans))