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>
using namespace std;

//repetition
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define rep(i, n) for(int i = 0; i < (int)(n); i++)

//container util
#define all(x) (x).begin(),(x).end()

//typedef
typedef long long ll;
typedef vector<int> VI;
typedef vector<VI> VVI;
typedef vector<ll> VLL;
typedef vector<VLL> VVLL;
typedef vector<string> VS;
typedef pair<int, int> PII;
typedef pair<ll, ll> PLL;


//conversion
inline int toInt(string s) {int v; istringstream sin(s);sin>>v;return v;}
template<class T> inline string toString(T x) {ostringstream sout;sout<<x;return sout.str();}


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