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>
#include <numeric>
#define rep(i,n) for (int i = 0; i < n; ++i)
#define rep1(i,n) for (int i = 1; i <= n; ++i)
template<typename T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
template<typename T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
using namespace std;
typedef long long ll;
typedef long double ld;
typedef pair<int,int> P;
const long long MOD = 1e9+7;
template<typename T> inline void add(T& a, T b) { a += b; if (a >= MOD) a -= MOD; }

int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);

  string s; cin >> s;
  vector<int> BI(26, -10);
  int n = s.size();
  rep(i, n) {
    if(i - BI[s[i] - 'a'] <= 2) {
      cout << BI[s[i] - 'a'] + 1 << " " << i + 1 << endl;
      return 0;
    }
    BI[s[i] - 'a'] = i;
  }
  cout << "-1 -1" << endl;
  return 0;
}