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 <iostream> // cout, endl, cin
// #include <string> // string, to_string, stoi
// #include <vector> // vector
// #include <algorithm> // min, max, swap, sort, reverse, lower_bound, upper_bound
// #include <utility> // pair, make_pair
// #include <tuple> // tuple, make_tuple
// #include <cstdint> // int64_t, int*_t
// #include <cstdio> // printf
// #include <map> // map
// #include <queue> // queue, priority_queue
// #include <set> // set
// #include <stack> // stack
// #include <deque> // deque
// #include <unordered_map> // unordered_map
// #include <unordered_set> // unordered_set
// #include <bitset> // bitset
// #include <climits>
// #include <cmath>
// #include <iomanip>
// #include <functional>
// #include <numeric>
// #include <random>
 
using namespace std;
    
#define int long long
#define pb push_back
#define eb emplace_back
// #define F first
// #define S second
#define FOR(i,a,b) for(int (i)=(a);(i)<(b);(i)++)
#define rep(i,n) FOR(i,0,n)
#define all(a) (a).begin(),(a).end()
#define rall(a) (a).rbegin(),(a).rend()
#define ve vector
#define vi vector<int>
#define vp vector<pair<int,int>>
#define vvi vector<vector<int>>
#define UNIQUE(a) sort(all(a)), a.erase(unique(all(a)), a.end())
 
template<typename T> using pq = priority_queue<T,vector<T>,greater<T>>; 
using ll = long long;
ll INF = LLONG_MAX / 2 - 100;
int IINF = INT_MAX / 4;
ll mod = 1e9 + 7;
int dx[] = {-1,0,1,0}, dy[] = {0,1,0,-1};
vector<ll> prime;

long double pi = 3.141592653589793238;
    
class fact {
public:
    long long fmod = 1e9+7;
    vector<long long> fac, finv, inv;
    fact (int n, long long Mod = 1e9+7) {
        fmod = Mod;
        fac = vector<long long>(n + 1, 0);
        finv = vector<long long>(n + 1, 0);
        inv = vector<long long>(n + 1, 0);
        fac[0] = fac[1] = 1;
        finv[0] = finv[1] = 1;
        inv[1] = 1; 
        for (int i = 2; i < n + 1; i++) {
            fac[i] = fac[i-1] * i % fmod;
            inv[i] = mod - inv[mod%i] * (mod/i) % mod;
            finv[i] = finv[i-1] * inv[i] % mod;
        }
    }
    ll nCr(ll n, ll r) {if(n < r) return 0; return fac[n] * finv[r] % fmod * finv[n-r] % fmod;}
    ll POW(ll a, ll b) {ll c = 1; while (b > 0) {if (b & 1) {c = a * c%fmod;}a = a * a%fmod; b >>= 1;}return c;}
    inline int operator [] (int i) {return fac[i];}
    ll DeBuG(ll n, ll r);
};
    
void DEBUG(vector<int> a) {for(int i=0;i<a.size();i++)cout<<a[i]<<" ";cout<<endl;}
void EMP(int x) {cout<<"!!!"<<x<<"!!!"<<endl;}
ll GCD(ll a, ll b) {ll c; while (b != 0) {c = a % b; a = b; b = c;}return a;}
ll LCM(ll a, ll b) {return (a / GCD(a, b)) * (b / GCD(a, b)) * GCD(a, b);}
ll POW(ll a, ll b) {ll c = 1; while (b > 0) {if (b & 1) {c = a * c%mod;}a = a * a%mod; b >>= 1;}return c;}
void PRI(ll n) {bool a[n + 1]; for (int i = 0; i < n + 1; i++) {a[i] = 1;}for (int i = 2; i < n + 1; i++) {if (a[i]) {prime.pb(i); ll b = i; while (b <= n) {a[b] = 0; b += i;}}}}
template <typename T> T chmin(T& a, T b) {if(a>b)a=b;return a;}
template <typename T> T chmax(T& a, T b) {if(a<b)a=b;return b;}
bool isSqrt(ll a) {return pow(sqrt(a),2) == a ? 1 : 0;}
void YesNo(bool a) {if (a) cout << "Yes"; else cout << "No"; cout << endl;}
void yesno(bool a) {if (a) cout << "yes"; else cout << "no"; cout << endl;}
void YESNO(bool a) {if (a) cout << "YES"; else cout << "NO"; cout << endl;}
double dis(int x1, int x2, int y1, int y2) {
    return sqrt((double)abs(x1-x2)*(double)abs(x1-x2)+(double)abs(y1-y2)*(double)abs(y1-y2));
}
int ceili(int x, int y) {
    if (x % y == 0) return x / y;
    else return x / y + 1;
}

void solve() {
    string s; cin >> s;
    if (s.size() == 2) {
        if (s[0] == s[1]) cout << 1 << " " << 2 << endl;
        else cout << -1 << " " << -1 << endl;
        return;
    }
    rep (i, (int)s.size()-1) {
        if (s[i] == s[i+1]) {cout << i+1 << " " << i+2 << endl; return;}
    }
    rep (i, (int)s.size()-2) {
        if (s[i] == s[i+2]) {cout << i+1 << " " << i+3 << endl; return;}
    }
    cout << -1 << " " << -1 << endl;
}
 
signed main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    solve();
}