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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>

#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define endl "\n"
#define rep(i, n) for(int i = 0; i < (int)(n); i++)
#define rrep(i, n) for(int i = ((int)(n)-1); i >= 0; i--)
#define all(x) (x).begin(),(x).end()

using namespace std;
using ll = long long;

typedef vector<int> vi;
typedef vector<vi> vvi;
typedef pair<int,int> P;
 
const int INF = 1e9;
//const ll INF = 1e18;
const double EPS = 1e-10;
const int MOD = 1e9+7;
const double PI = acos(-1.0);

int main()
{
	string s;
	cin >> s;
	if(s.size() == 2 && s[0] == s[1]){
		cout << "1 2" << endl;
		return 0;
	}
	if(s.size() >= 3 && s[0] == s[1]){
		cout << "1 3" << endl;
		return 0;
	}
	for(int i = 2; i < s.size(); i++){
		if(s[i-1] == s[i]){
			cout << i-1 << ' ' << i+1 << endl;
			return 0;
		}
	}
	for(int i = 1; i < s.size()-1; i++){
		if(s[i-1] == s[i+1]){
			cout << i << ' ' << i+2 << endl;
			return 0;
		}
	}
	cout << "-1 -1" << endl;
	return 0;
}