//solutia cu deque, pentru ca e o structura de date utila uneori
#include <fstream>
//#include <iostream>
#include <deque>
using namespace std;

int v[500003];
deque <int> d;
class InParser
{
private:
    FILE *fin;
    char *buff;
    int sp;

    char read_ch()
    {
        ++sp;
        if (sp == 4096)
        {
            sp = 0;
            fread(buff, 1, 4096, fin);
        }
        return buff[sp];
    }

public:
    InParser(const char* nume)
    {
        fin = fopen(nume, "r");
        buff = new char[4096]();
        sp = 4095;
    }

    InParser& operator >> (int &n)
    {
        char c;
        while (!isdigit(c = read_ch()) && c != '-');
        int sgn = 1;
        if (c == '-')
        {
            n = 0;
            sgn = -1;
        }
        else
        {
            n = c - '0';
        }
        while (isdigit(c = read_ch()))
        {
            n = 10 * n + c - '0';
        }
        n *= sgn;
        return *this;
    }

    InParser& operator >> (long long &n)
    {
        char c;
        n = 0;
        while (!isdigit(c = read_ch()) && c != '-');
        long long sgn = 1;
        if (c == '-')
        {
            n = 0;
            sgn = -1;
        }
        else
        {
            n = c - '0';
        }
        while (isdigit(c = read_ch()))
        {
            n = 10 * n + c - '0';
        }
        n *= sgn;
        return *this;
    }
};
signed main()
{
    ofstream cout("secventa.out");
    InParser cin("secventa.in");
    int n, k, i;
    cin >> n >> k;
    int ans = -30003, a;
    for (i = 1; i <= n; i++)
    {
        cin >> v[i];
        while (!d.empty() and d.front() <= i - k)
            d.pop_front();
        while (!d.empty() and v[d.back()] > v[i])
            d.pop_back();
        d.push_back(i);
        if (v[d.front()] > ans and i >= k)
        {
            ans = v[d.front()];
            a = i - k + 1;
        }
    }
    cout << a << " " << a + k - 1 << " " << ans;
}