Problem: You are given n points with integer coordinates on the plane. Points are given in a way such that there is no triangle, formed by any three of these n points, which area exceeds S.

Alyona tried to construct a triangle with integer coordinates, which contains all n points and which area doesn't exceed 4S, but, by obvious reason, had no success in that. Please help Alyona construct such triangle. Please note that vertices of resulting triangle are not necessarily chosen from n given points.

Input

In the first line of the input two integers n and S (3 ≤ n ≤ 5000, 1 ≤ S ≤ 1018) are given — the number of points given and the upper bound value of any triangle's area, formed by any three of given n points.

The next n lines describes given points: ith of them consists of two integers xi and yi ( - 108 ≤ xi, yi ≤ 108) — coordinates of ith point.

It is guaranteed that there is at least one triple of points not lying on the same line.

Output

Print the coordinates of three points — vertices of a triangle which contains all n points and which area doesn't exceed 4S.

Coordinates of every triangle's vertex should be printed on a separate line, every coordinate pair should be separated by a single space. Coordinates should be an integers not exceeding 109 by absolute value.

It is guaranteed that there is at least one desired triangle. If there is more than one answer, print any of them.

Example

Input

4 1
0 0
1 0
0 1
1 1


Output

-1 0
2 0
0 2

Note

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Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
void io() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cout.precision(15);
}
const long double EPS = 1e-18;
struct PT {
  long double x, y;
  bool operator<(const PT &rhs) const {
    return make_pair(y, x) < make_pair(rhs.y, rhs.x);
  }
  bool operator==(const PT &rhs) const {
    return make_pair(y, x) == make_pair(rhs.y, rhs.x);
  }
  PT() {}
  PT(long double x, long double y) : x(x), y(y) {}
  PT(const PT &p) : x(p.x), y(p.y) {}
  PT operator+(const PT &p) const { return PT(x + p.x, y + p.y); }
  PT operator-(const PT &p) const { return PT(x - p.x, y - p.y); }
  PT operator*(long double c) const { return PT(x * c, y * c); }
  PT operator/(long double c) const { return PT(x / c, y / c); }
};
long double cross(PT p, PT q) { return p.x * q.y - p.y * q.x; }
long double area2(PT a, PT b, PT c) {
  return cross(a, b) + cross(b, c) + cross(c, a);
}
bool between(const PT &a, const PT &b, const PT &c) {
  return (fabs(area2(a, b, c)) < EPS && (a.x - b.x) * (c.x - b.x) <= 0 &&
          (a.y - b.y) * (c.y - b.y) <= 0);
}
void ConvexHull(vector<PT> &pts) {
  if (pts.size() < 3) return;
  sort(pts.begin(), pts.end());
  pts.erase(unique(pts.begin(), pts.end()), pts.end());
  vector<PT> up, dn;
  for (int i = 0; i < pts.size(); i++) {
    while (up.size() > 1 && area2(up[up.size() - 2], up.back(), pts[i]) >= 0)
      up.pop_back();
    while (dn.size() > 1 && area2(dn[dn.size() - 2], dn.back(), pts[i]) <= 0)
      dn.pop_back();
    up.push_back(pts[i]);
    dn.push_back(pts[i]);
  }
  pts = dn;
  for (int i = (int)up.size() - 2; i >= 1; i--) pts.push_back(up[i]);
  if (pts.size() <= 2) return;
  dn.clear();
  dn.push_back(pts[0]);
  dn.push_back(pts[1]);
  for (int i = 2; i < pts.size(); i++) {
    if (between(dn[dn.size() - 2], dn[dn.size() - 1], pts[i])) dn.pop_back();
    dn.push_back(pts[i]);
  }
  if (dn.size() >= 3 && between(dn.back(), dn[0], dn[1])) {
    dn[0] = dn.back();
    dn.pop_back();
  }
  pts = dn;
}
long double dot(PT p, PT q) { return p.x * q.x + p.y * q.y; }
long double dist2(PT p, PT q) { return dot(p - q, p - q); }
ostream &operator<<(ostream &os, const PT &p) {
  os << "(" << p.x << "," << p.y << ")";
}
PT RotateCCW90(PT p) { return PT(-p.y, p.x); }
PT RotateCW90(PT p) { return PT(p.y, -p.x); }
PT RotateCCW(PT p, long double t) {
  return PT(p.x * cos(t) - p.y * sin(t), p.x * sin(t) + p.y * cos(t));
}
PT ProjectPointLine(PT a, PT b, PT c) {
  return a + (b - a) * dot(c - a, b - a) / dot(b - a, b - a);
}
PT ProjectPointSegment(PT a, PT b, PT c) {
  long double r = dot(b - a, b - a);
  if (fabs(r) < EPS) return a;
  r = dot(c - a, b - a) / r;
  if (r < 0) return a;
  if (r > 1) return b;
  return a + (b - a) * r;
}
long double DistancePointSegment(PT a, PT b, PT c) {
  return sqrt(dist2(c, ProjectPointSegment(a, b, c)));
}
bool PointInPolygon(const vector<PT> &p, PT q) {
  bool c = 0;
  for (int i = 0; i < p.size(); i++) {
    int j = (i + 1) % p.size();
    if ((p[i].y <= q.y && q.y < p[j].y || p[j].y <= q.y && q.y < p[i].y) &&
        q.x < p[i].x + (p[j].x - p[i].x) * (q.y - p[i].y) / (p[j].y - p[i].y))
      c = !c;
  }
  return c;
}
bool PointOnPolygon(const vector<PT> &p, PT q) {
  for (int i = 0; i < p.size(); i++)
    if (dist2(ProjectPointSegment(p[i], p[(i + 1) % p.size()], q), q) < EPS)
      return true;
  return false;
}
double tri_ar(PT a, PT b, PT c) {
  double ret = 0.0;
  ret = fabs(cross(c - a, b - a));
  ret = ret * 0.5;
  return ret;
}
vector<PT> pts;
vector<PT> get_max_triangle(vector<PT> &pts) {
  vector<PT> ret;
  int ai = 0, bi = 1, ci = 2;
  double maxar = 0.0;
  int n = pts.size();
  vector<PT> an_pt(3);
  for (int i = 0; i < n; i++) {
    ai = i;
    if (bi == ai) bi = (bi + 1) % n;
    if (ci == bi) ci = (ci + 1) % n;
    while (1) {
      while (1) {
        if (tri_ar(pts[ai], pts[bi], pts[ci]) <=
            tri_ar(pts[ai], pts[bi], pts[(ci + 1) % n]))
          ci = (ci + 1) % n;
        else
          break;
      }
      if (tri_ar(pts[ai], pts[bi], pts[ci]) <=
          tri_ar(pts[ai], pts[(bi + 1) % n], pts[ci])) {
        bi = (bi + 1) % n;
      } else
        break;
    }
    if ((maxar + EPS) < tri_ar(pts[ai], pts[bi], pts[ci])) {
      maxar = tri_ar(pts[ai], pts[bi], pts[ci]);
      an_pt[0] = pts[ai];
      an_pt[1] = pts[bi];
      an_pt[2] = pts[ci];
    }
  }
  return an_pt;
}
int main(int argc, char *argv[]) {
  io();
  int n;
  cin >> n;
  double S, x, y;
  cin >> S;
  for (int i = 0; i < n; i++) {
    cin >> x >> y;
    pts.push_back(PT{x, y});
  }
  ConvexHull(pts);
  vector<PT> maxtri = get_max_triangle(pts);
  PT a = maxtri[1] + maxtri[2] - maxtri[0];
  PT b = maxtri[0] - maxtri[1] + maxtri[2];
  PT c = maxtri[0] + maxtri[1] - maxtri[2];
  cout << a.x << " " << a.y << '\n';
  cout << b.x << " " << b.y << '\n';
  cout << c.x << " " << c.y << '\n';
  return 0;
}