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: Time Limit Exceeded)

#include <bits/stdc++.h>
#define ll long long
#define pb push_back
#define ep emplace_back
#define st first
#define sc second

using namespace std;

const ll mod = 998244353;
const int e = 1e6 + 69;
const int base = 311;

// dung igt thay cho gt luc tinh giai thua
// 3
// cach tao hoan vi tu n so : n*(n+1)/2;

ll a[e];
ll dp[e][27];

int main()
{
    string s;
    cin >> s;
    s = ' ' + s;
    for(int i=1; i<s.size(); i++)
    {
        for(int j=0; j<26; j++)
        {
            dp[i][j] = dp[i-1][j];
            if(s[i] == j + 'a') dp[i][j] ++;
        }
    }
    for(int c=0; c<26; c++)
    {
        for(int i=1; i<s.size() - 1; i++)
        {
            for(int j=i+1; j<s.size(); j++)
            {
                ll cur = dp[j][c] - dp[i-1][c];
                if(cur*2 > j - i + 1)
                {
                    cout << i <<" "<<j;
                    return 0;
                }
            }
        }
    }
    cout << - 1 <<" "<< -1;
}