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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
///Welcome to Nasif's Code
#define bug printf("bug\n");
#define bug2(var) cout<<#var<<" "<<var<<endl;
#define co(q) cout<<q<<endl;
#define all(q) (q).begin(),(q).end()
typedef long long int ll;
typedef unsigned long long int ull;
const int MOD = (int)1e9+7;
const int MAX = 1e6;
#define pi acos(-1)
#define inf 1000000000000000LL
#define FastRead    ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
int freq[26][MAX];
int main()
{
    FastRead
    //freopen("output.txt", "w", stdout);
    string s;
    cin>>s;
    for(int i=0; i<s.length(); i++)
    {
        if(i)
        for(int j=0; j<26; j++){
            if(freq[j][i-1])
            freq[j][i]=freq[j][i-1];
        }
        freq[s[i]-'a'][i]++;
    }
    int len=s.length();
    for(int i=0;i<26;i++){
        for(int j=0;j<len;j++){
            if(freq[i][j]>2){
                cout<<(j-2)+1<<" "<<j+1<<endl;
                return 0;
            }
        }
    }


  cout<<"-1 -1"<<endl;
    return 0;
}