/* pku_半平面交算法的一个应用(两圆覆盖凸多边形,只能是凸多边形）
 具体做法是，把多边形的每条边向内平移r单位长度，用这些线段所在直线和原多边形作半平面交，得到的区域就是半径为r的圆放入多边形的可行域。可以证明这个区域一定是凸的，或者退化为一条线段，或一个点。那么，我们就可以在这个区域上求最远点对啦。
 */
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<iostream>
#include<algorithm>
using namespace std;
#define maxn 110
#define eps 1e-8
#define inf 0xffffff

struct Nodes {
    double x, y;
};

struct Set {
    int nn;
    Nodes p[300];
};
int n;
Set src;

double cross(Nodes p1, Nodes p2, Nodes p3) {
    return ((p2.x - p1.x)*(p3.y - p1.y)-(p2.y - p1.y)*(p3.x - p1.x));
}

double sqr(double x) {
    return x*x;
}

double dis(Nodes p1, Nodes p2) {
    return sqrt(sqr(p1.x - p2.x) + sqr(p1.y - p2.y));
}

void Getline(Nodes p1, Nodes p2, double &a, double &b, double &c) {
    a = p2.y - p1.y;
    b = p1.x - p2.x;
    c = p1.y * (p2.x - p1.x) - p1.x * (p2.y - p1.y);
}

Nodes get_crossp(double a1, double b1, double c1, double a2, double b2, double c2) {
    Nodes ret;
    ret.x = -(c1 * b2 - b1 * c2) / (a1 * b2 - a2 * b1);
    ret.y = -(a1 * c2 - a2 * c1) / (a1 * b2 - a2 * b1);
    return ret;
}

int isequal(Nodes p1, Nodes p2) {
    if (fabs(p1.x - p2.x) < eps && fabs(p1.y - p2.y) < eps)return 1;
    return 0;
}

Set cut(Nodes p1, Nodes p2, Set s) {
    int i, j, k;
    double t1, t2, a1, b1, c1, a2, b2, c2;
    Nodes crossp;
    Set ret;
    ret.nn = 0;
    for (i = 0; i < s.nn; i++) {
        t1 = cross(p1, p2, s.p[i]);
        t2 = cross(p1, p2, s.p[i + 1]);
        if (t1 < eps && t2 < eps) {
            ret.p[ret.nn++] = s.p[i];
            ret.p[ret.nn++] = s.p[i + 1];
        } else if (t1 > eps && t2 > eps) {
            continue;
        } else {
            Getline(p1, p2, a1, b1, c1);
            Getline(s.p[i], s.p[i + 1], a2, b2, c2);
            crossp = get_crossp(a1, b1, c1, a2, b2, c2);
            if (t1 < eps) {
                ret.p[ret.nn++] = s.p[i];
                ret.p[ret.nn++] = crossp;
            } else {
                ret.p[ret.nn++] = crossp;
                ret.p[ret.nn++] = s.p[i + 1];
            }
        }
    }
    if (ret.nn == 0)return ret;
    j = 1;
    for (i = 1; i < ret.nn; i++) {
        if (isequal(ret.p[i - 1], ret.p[i]))continue;
        ret.p[j++] = ret.p[i];
    }
    ret.nn = j;
    if (ret.nn != 1 && isequal(ret.p[ret.nn - 1], ret.p[0]))ret.nn--;
    ret.p[ret.nn] = ret.p[0];
    return ret;
}

int main() {
    int i, j, k, r;
    Set s;
    Nodes p1, p2;
    double dx, dy;
    double len, maxdis;
    while (scanf("%d%d", &n, &r) != EOF) {
        src.nn = n;
        for (i = 0; i < src.nn; i++) {
            scanf("%lf%lf", &src.p[i].x, &src.p[i].y);
        }
        src.p[src.nn] = src.p[0];
        s = src;
        for (i = 0; i < src.nn; i++) {
            dx = src.p[i + 1].x - src.p[i].x;
            dy = src.p[i + 1].y - src.p[i].y;
            len = dis(src.p[i + 1], src.p[i]);
            p1.x = src.p[i].x + dy * r / len;
            p1.y = src.p[i].y - dx * r / len;
            p2.x = src.p[i + 1].x + dy * r / len;
            p2.y = src.p[i + 1].y - dx * r / len;
            s = cut(p1, p2, s);
        }
        maxdis = -inf * 1.0;
        for (i = 0; i < s.nn; i++) {
            for (j = i; j < s.nn; j++) {
                len = dis(s.p[i], s.p[j]);
                if (len > maxdis) {
                    maxdis = len;
                    p1 = s.p[i];
                    p2 = s.p[j];
                }
            }
        }
        printf("%.4lf %.4lf %.4lf %.4lf\n", p1.x, p1.y, p2.x, p2.y);
    }
    return 0;
}