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>
#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
const int INF = 0x3f3f3f3f;
const long long LINF = 0x3f3f3f3f3f3f3f3fLL;
const int MOD = 1e9 + 7;
const double pi = acos(-1.0);
const double eps = 1e-6;
const int maxn = 5e3 + 5;
const int maxm = 1e6 + 5;
const int dx[] = {-1, 0, 1, 0, -1, -1, 1, 1};
const int dy[] = {0, 1, 0, -1, 1, -1, 1, -1};
inline int scan(int &a) { return scanf("%d", &a); }
inline int scan(int &a, int &b) { return scanf("%d%d", &a, &b); }
inline int scan(int &a, int &b, int &c) { return scanf("%d%d%d", &a, &b, &c); }
inline int scan(long long &a) { return scanf("%I64d", &a); }
inline int scan(long long &a, long long &b) {
  return scanf("%I64d%I64d", &a, &b);
}
inline int scan(long long &a, long long &b, long long &c) {
  return scanf("%I64d%I64d%I64d", &a, &b, &c);
}
inline int scan(double &a) { return scanf("%lf", &a); }
inline int scan(double &a, double &b) { return scanf("%lf%lf", &a, &b); }
inline int scan(double &a, double &b, double &c) {
  return scanf("%lf%lf%lf", &a, &b, &c);
}
inline int scan(char &a) { return scanf("%c", &a); }
inline int scan(char *a) { return scanf("%s", a); }
template <class T>
inline void mem(T &A, int x) {
  memset(A, x, sizeof(A));
}
template <class T0, class T1>
inline void mem(T0 &A0, T1 &A1, int x) {
  mem(A0, x), mem(A1, x);
}
template <class T0, class T1, class T2>
inline void mem(T0 &A0, T1 &A1, T2 &A2, int x) {
  mem(A0, x), mem(A1, x), mem(A2, x);
}
template <class T0, class T1, class T2, class T3>
inline void mem(T0 &A0, T1 &A1, T2 &A2, T3 &A3, int x) {
  mem(A0, x), mem(A1, x), mem(A2, x), mem(A3, x);
}
template <class T0, class T1, class T2, class T3, class T4>
inline void mem(T0 &A0, T1 &A1, T2 &A2, T3 &A3, T4 &A4, int x) {
  mem(A0, x), mem(A1, x), mem(A2, x), mem(A3, x), mem(A4, x);
}
template <class T0, class T1, class T2, class T3, class T4, class T5>
inline void mem(T0 &A0, T1 &A1, T2 &A2, T3 &A3, T4 &A4, T5 &A5, int x) {
  mem(A0, x), mem(A1, x), mem(A2, x), mem(A3, x), mem(A4, x), mem(A5, x);
}
template <class T0, class T1, class T2, class T3, class T4, class T5, class T6>
inline void mem(T0 &A0, T1 &A1, T2 &A2, T3 &A3, T4 &A4, T5 &A5, T6 &A6, int x) {
  mem(A0, x), mem(A1, x), mem(A2, x), mem(A3, x), mem(A4, x), mem(A5, x),
      mem(A6, x);
}
template <class T>
inline T min(T a, T b, T c) {
  return min(min(a, b), c);
}
template <class T>
inline T max(T a, T b, T c) {
  return max(max(a, b), c);
}
template <class T>
inline T min(T a, T b, T c, T d) {
  return min(min(a, b), min(c, d));
}
template <class T>
inline T max(T a, T b, T c, T d) {
  return max(max(a, b), max(c, d));
}
template <class T>
inline T min(T a, T b, T c, T d, T e) {
  return min(min(min(a, b), min(c, d)), e);
}
template <class T>
inline T max(T a, T b, T c, T d, T e) {
  return max(max(max(a, b), max(c, d)), e);
}
template <class T>
inline void dbg(T a[], int n) {
  for (int i = 0; i < n; i++) cout << a[i] << (i == n - 1 ? "\n" : " ");
}
template <class T>
inline void dbg(T a) {
  cout << a << endl;
}
template <class T>
inline void dbg(T a[][maxn], int n, int m) {
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++) cout << a[i][j] << (j == m - 1 ? "\n" : " ");
}
inline int Sgn(double x) {
  if (fabs(x) < eps)
    return 0;
  else if (x > 0)
    return 1;
  else
    return -1;
}
struct Point {
  double x, y;
  Point() {}
  Point(double _x, double _y) {
    x = _x;
    y = _y;
  }
  Point operator-(const Point &b) const { return Point(x - b.x, y - b.y); }
  double operator^(const Point &b) const { return x * b.y - y * b.x; }
  double operator*(const Point &b) const { return x * b.x + y * b.y; }
  void TransXY(double B) {
    double tx = x, ty = y;
    x = tx * cos(B) - ty * sin(B);
    y = tx * sin(B) + ty * cos(B);
  }
} p[maxn];
double gao(Point a, Point b, Point c) {
  double res = 0.0;
  res += ((a ^ b) + (b ^ c) + (c ^ a)) / 2.0;
  return fabs(res);
}
int main() {
  int n;
  long long s;
  scan(n);
  scan(s);
  for (int i = 0; i < n; i++) scan(p[i].x, p[i].y);
  int a = 0, b = 1, c = 2;
  double ans = gao(p[a], p[b], p[c]);
  int flag = 1;
  while (flag) {
    flag = 0;
    for (int i = 0; i < n; i++) {
      double t;
      t = gao(p[a], p[b], p[i]);
      if (t > ans) {
        ans = t;
        c = i;
        flag = 1;
      }
      t = gao(p[a], p[i], p[c]);
      if (t > ans) {
        ans = t;
        b = i;
        flag = 1;
      }
      t = gao(p[i], p[b], p[c]);
      if (t > ans) {
        ans = t;
        a = i;
        flag = 1;
      }
    }
  }
  printf("%.0f %.0f\n", p[b].x + p[c].x - p[a].x, p[b].y + p[c].y - p[a].y);
  printf("%.0f %.0f\n", p[a].x + p[c].x - p[b].x, p[a].y + p[c].y - p[b].y);
  printf("%.0f %.0f\n", p[b].x + p[a].x - p[c].x, p[b].y + p[a].y - p[c].y);
  return 0;
}