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;
struct Point {
  int x, y;
  Point() {}
  Point(int x, int y) : x(x), y(y) {}
  double norm() { return sqrt(1LL * x * x + 1LL * y * y); }
};
Point operator+(const Point &A, const Point &B) {
  return Point(A.x + B.x, A.y + B.y);
}
Point operator-(const Point &A, const Point &B) {
  return Point(A.x - B.x, A.y - B.y);
}
bool operator<(const Point &A, const Point &B) {
  return make_pair(A.x, A.y) < make_pair(B.x, B.y);
}
double dist(const Point &A, const Point &B) {
  Point AB = B - A;
  return AB.norm();
}
int n, chN;
long long S;
Point p[5005];
long long crossProduct(Point A, Point B, Point C) {
  Point AB = B - A;
  Point AC = C - A;
  return 1LL * AB.x * AC.y - 1LL * AB.y * AC.x;
}
long long triangleArea2(Point A, Point B, Point C) {
  return abs(crossProduct(A, B, C));
}
bool Comp(Point A, Point B) {
  long long cp = crossProduct(p[1], A, B);
  return cp > 0;
}
int nextIdx(int idx) {
  if (++idx == chN) return 0;
  return idx;
}
int main() {
  scanf("%d %I64d", &n, &S);
  for (int i = 1; i <= n; ++i) scanf("%d %d", &p[i].x, &p[i].y);
  for (int i = 2; i <= n; ++i) {
    if (p[i] < p[1]) swap(p[i], p[1]);
  }
  sort(p + 2, p + n + 1, Comp);
  vector<Point> convexHull;
  convexHull.push_back(p[1]);
  for (int i = 2; i <= n; ++i) {
    Point A = p[i];
    while (1) {
      if ((int)convexHull.size() < 2) break;
      Point P1 = convexHull[(int)convexHull.size() - 2];
      Point P2 = convexHull.back();
      long long cp = crossProduct(P1, P2, A);
      if (cp <= 0)
        convexHull.pop_back();
      else
        break;
    }
    convexHull.push_back(A);
  }
  int i = 0, j = 1, k = 2;
  long long maxArea = 0;
  int idx1, idx2, idx3;
  chN = (int)convexHull.size();
  vector<Point> &ch = convexHull;
  for (i = 0; i < chN; ++i) {
    if (i == j) j = nextIdx(j);
    if (j == k) k = nextIdx(k);
    while (1) {
      long long area = triangleArea2(ch[i], ch[j], ch[k]);
      while (triangleArea2(ch[i], ch[j], ch[nextIdx(k)]) >= area) {
        k = nextIdx(k);
        area = triangleArea2(ch[i], ch[j], ch[k]);
      }
      if (area > maxArea) {
        maxArea = area;
        idx1 = i;
        idx2 = j;
        idx3 = k;
      }
      if (triangleArea2(ch[i], ch[nextIdx(j)], ch[k]) < area) break;
      j = nextIdx(j);
    }
  }
  Point P1 = ch[idx1];
  Point P2 = ch[idx2];
  Point P3 = ch[idx3];
  Point S1 = P2 + P3 - P1;
  Point S2 = P1 + P3 - P2;
  Point S3 = P1 + P2 - P3;
  printf("%d %d\n", S1.x, S1.y);
  printf("%d %d\n", S2.x, S2.y);
  printf("%d %d\n", S3.x, S3.y);
  return 0;
}