#include <stdio.h>
#define NMax 500005

const char IN[]="secventa.in",OUT[]="secventa.out";

int N,K;
int x,y,Max=-1;
int a[NMax],D[NMax],first,last;

void solve()
{
	using namespace std;
	int i;
	for (i=0;i<N;i++)
	{
		while (first<=last && (a[i]<a[D[last]] || i-D[last]>=K))
			last--;
		D[++last]=i;
		while (first<=last && i-D[first]>=K)
			first++;
		if (i==K-1)
			Max=a[D[first]],x=0,y=K-1;
		if (i>=K && a[D[first]]>Max)
			Max= a[D[first]],x=i-K+1,y=x+K-1;
	}
}

int main()
{
	int i;
	freopen(IN,"r",stdin);
	scanf("%d%d",&N,&K);
	for (i=0;i<N;i++)
		scanf("%d",a+i);
	fclose(stdin);
	solve();
	
	freopen(OUT,"w",stdout);
	printf("%d %d %d",x+1,y+1,Max);
	fclose(stdout);
	
	return 0;
}