//ALEXANDRU MICLEA

#include <vector>
#include <algorithm>
#include <string>
#include <string.h>
#include <cstring>
#include <queue>
#include <map>
#include <set>
#include <unordered_map>
#include <time.h>
#include <iomanip>
#include <deque>
#include <math.h>
#include <cmath>
#include <assert.h>
#include <stack>
#include <bitset>
#include <random>
#include <chrono>
#include <assert.h>
#include <iostream>

using namespace std;
using ll = long long;

#define fast_cin() 	ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)

//VARIABLES

deque <int> q;
int ans = 0;
int v[500005];

//FUNCTIONS



//MAIN
int main() {

	#ifdef INFOARENA
        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;
            }
        }      ("secventa.in", "r", stdin);
		freopen("secventa.out", "w", stdout);
	#endif

	int n, k; cin >> n >> k;
	int x = 0, y = 0;
	int ans = -500005;

	for (int i = 1; i <= n; i++) cin >> v[i];

	for (int i = 1; i <= n; i++) {
		while (!q.empty() && v[q.back()] > v[i]) q.pop_back();
		q.push_back(i);
		while (!q.empty() && q.front() + k <= i) q.pop_front();

		if (!(k > i) && ans < v[q.front()]) {
			ans = v[q.front()];
			x = i - k + 1;
			y = i;
		}
	}

	cout << x << ' ' << y << ' ' << ans;

	return 0;
}