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: Wrong Answer)

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef pair<int,int> pp;

int N;
const int MAX_N = 100010;
const int MAX_M = 26;
const int INF = 1e9;

char a[MAX_N];
int min_v[MAX_M], min_i[MAX_M], cnt[MAX_M];

pp solve(int L,int R){
    if(L+1>=R)return pp(-1,-1);
    int mid = (L+R)/2;
    pp ret = solve(L, mid);
    if(ret.first>=0)return ret;
    ret = solve(mid+1, R);
    if(ret.first>=0)return ret;

    for(int i=L;i<=mid;i++){
        int x = a[i]-'a';
        if(min_v[x] > cnt[x]*2-i+1){
            min_v[x] = cnt[x]*2-i+1;
            min_i[x] = i;
        }
        cnt[x]++;
    }
    for(int i=mid+1;i<R;i++){
        int x = a[i]-'a';
        cnt[x]++;
        if(2*cnt[x]-i > min_v[x]){
            return pp(min_i[x], i);
        }
    }

    for(int i=L;i<=mid;i++){
        int x = a[i]-'a';
        min_v[x] = INF;
        cnt[x] = 0;
    }
    return pp(-1,-1);
}

int main(){
    scanf("%s",a+1);
    N = strlen(a+1);

    for(int i=0;i<MAX_M;i++){
        min_v[i] = INF;
    }
    pp ans = solve(1,N+1);

    printf("%d %d\n",ans.first, ans.second);
    return 0;
}