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;
#define rep(i,a,b) for(int i=a;i<b;i++)
#define rrep(i,a,b) for(int i=a;i>=b;i--)
#define erep(i,a,n) for(int i = a;i<=n;i++)
typedef long long ll;
#define int long long
#define vint vector<int>
#define vvint vector<vector<int>>
#define vstring vector<string>
#define vdouble vector<double>
#define vll vector<ll>:
#define vbool vector<bool>
#define INF 1101010101010101010
#define MOD 1000000007
#define P pair<int,int>
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a = b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a = b; return 1; } return 0; }

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