#include <iostream>
#include <fstream>
#include <cstdlib>
#include <ctime>
#include <deque>

using namespace std;
ifstream in("secventa.in");
ofstream out("secventa.out");

#define ll long long
#define ull unsigned long long
#define ui unsigned int
#define pb push_back
const int NMax = 5e5 + 5;
const int strMax = 7*NMax + 50;
const short inf = 31e3 + 5;

int N,K;
char str[strMax],*p = str;
int v[NMax];
deque<int> deq;
// str,*p - pentru parsare;
// v - array din input;
// deq - double ended queue pentru obtinerea bazelor;
// se obtine minimul pentru fiecare secventa de lungime K,
// ca la problema deque din arhiva educationala;
// Este necesara doar verificarea celor de lungime K, deoarece
// minimul nu poate creste prin marirea secventei;

int getNumber();
// functie pentru parsare;

int main() {
    in.get(str,strMax,'&');

    N = getNumber();
    K = getNumber();
    for (int i=1;i < K;++i) {
        v[i] = getNumber();

        while (deq.size() && v[deq.back()] >= v[i]) {
            deq.pop_back();
        }
        deq.push_back(i);
    }

    int mx = -inf;
    int st = -1,dr = -1;
    for (int i=K;i <= N;++i) {
        v[i] = getNumber();

        if (deq.front() == i-K) {
            deq.pop_front();
        }

        while (deq.size() && v[deq.back()] >= v[i]) {
            deq.pop_back();
        }
        deq.push_back(i);

        if (v[deq.front()] > mx) {
            mx = v[deq.front()];
            st = i-K+1;
            dr = i;
        }
    }
    out<<st<<' '<<dr<<' '<<mx<<'\n';

    return 0;
}

int getNumber() {
    bool neg = false;
    while (!('0' <= *p && *p <= '9')) {
        if (*p == '-') {
            neg = true;
        }
        ++p;
    }

    int ans = 0;
    while ('0' <= *p && *p <= '9') {
        ans = ans * 10 + *p++ - '0';
    }

    return (neg) ? -ans : ans;
}