#include <iostream>
#include <fstream>
#include <math.h>
using namespace std;
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;
	}
};
InParser f("secventa.in");
ofstream g("secventa.out");
int mprec[500005][23];
 int n,k,v[500000];
 int l2k;
void precalc(int *v,int n)
{
    int p2=2,logp2=1;
    for(int i=0;i<n;i++)
        mprec[i][0]=v[i];
    while(p2<=n){
        for(int i=0;i<n-p2+1;i++)
            {
                mprec[i][logp2]=min(mprec[i][logp2-1],mprec[i+p2/2][logp2-1]);
            }
        logp2++;
        p2=p2*2;
    }
}
int vmin(int a,int b)
{
  return min(mprec[a][l2k],mprec[b-(1<<l2k)+1][l2k]);
}
int main()
{

  f>>n>>k;
  for(int i=0;i<n;i++)
    f>>v[i];
  precalc(v,n);

  l2k=log2(k);
    int mx=vmin(0,k-1);
    int imax=0;
    for(int i=1;i<n-k+1;i++)
    {
        int minim=vmin(i,i+k-1);
        if(minim>mx)
            {
                mx=minim;
                imax=i;
            }

    }
    g<<imax+1<<" "<<imax+k<<" "<<mx;

}