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;
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast")
#pragma GCC optimize(3)
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC target("sse3", "sse2", "sse")
#pragma GCC target("avx", "sse4", "sse4.1", "sse4.2", "ssse3")
#pragma GCC target("f16c")
#pragma GCC optimize("inline", "fast-math", "unroll-loops", \
                     "no-stack-protector")
#pragma GCC push_options
#pragma GCC pop_options
#pragma GCC reset_options
#pragma GCC optimize("O3")
const double eps = 1e-9;
const int INFMEM = 63;
const int INF = 1061109567;
const long long LINF = 4557430888798830399LL;
const double DINF = numeric_limits<double>::infinity();
const long long MOD = 1000000007;
const int dx[8] = {1, 0, -1, 0, 1, 1, -1, -1};
const int dy[8] = {0, 1, 0, -1, 1, -1, 1, -1};
const double PI = 3.141592653589793;
inline void fastll(long long &input_number) {
  input_number = 0;
  int ch = getchar_unlocked();
  int sign = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') sign = -1;
    ch = getchar_unlocked();
  }
  while (ch >= '0' && ch <= '9') {
    input_number = (input_number << 3) + (input_number << 1) + ch - '0';
    ch = getchar_unlocked();
  }
  input_number *= sign;
}
inline void open(string a) {
  freopen((a + ".in").c_str(), "r", stdin);
  freopen((a + ".out").c_str(), "w", stdout);
}
inline void fasterios() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
long long t, n, s;
struct point {
  long long x, y;
  bool operator<(const point &other) const {
    if (x != other.x) return x < other.x;
    return y < other.y;
  }
};
struct triangle {
  point a, b, c;
  long long area;
  bool operator<(const triangle &other) const { return area < other.area; }
  triangle() {
    a = {0, 0};
    b = {0, 0};
    c = {0, 0};
    area = 0;
  }
  triangle(point aa, point bb, point cc) {
    a = aa;
    b = bb;
    c = cc;
    area = (a.x * b.y + b.x * c.y + c.x * a.y) -
           (a.y * b.x + b.y * c.x + c.y * a.x);
    area = abs(area);
  }
};
point pivot;
point isi[5005];
vector<point> ch;
long long solve(point p, point q, point r) {
  long long cross = (r.x - q.x) * (p.y - q.y) - (r.y - q.y) * (p.x - q.x);
  if (cross == 0)
    return 0;
  else if (cross < 0)
    return -1;
  else
    return 1;
}
long long dist(point a, point b) {
  return (a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y);
}
bool cmp(point a, point b) {
  long long dir = solve(pivot, a, b);
  if (dir == 0) return dist(pivot, a) < dist(pivot, b);
  return (dir == -1);
}
void print(triangle other) {
  cout << "This triangle has " << '\n';
  cout << other.a.x << " " << other.a.y << '\n';
  cout << other.b.x << " " << other.b.y << '\n';
  cout << other.c.x << " " << other.c.y << '\n';
  cout << "With area " << other.area / 2;
  if (other.area & 1) cout << ".5";
  cout << '\n';
}
void printans(triangle ans) {
  point aa = {-ans.a.x + ans.b.x + ans.c.x, -ans.a.y + ans.b.y + ans.c.y};
  point bb = {ans.a.x - ans.b.x + ans.c.x, ans.a.y - ans.b.y + ans.c.y};
  point cc = {ans.a.x + ans.b.x - ans.c.x, ans.a.y + ans.b.y - ans.c.y};
  cout << aa.x << " " << aa.y << '\n';
  cout << bb.x << " " << bb.y << '\n';
  cout << cc.x << " " << cc.y << '\n';
}
int main() {
  cin >> n >> s;
  for (int i = 1; i <= n; i++) {
    cin >> isi[i].x >> isi[i].y;
    if (i == 1) pivot = isi[i];
    pivot = min(pivot, isi[i]);
  }
  sort(isi + 1, isi + 1 + n, cmp);
  ch.clear();
  for (int i = 1; i <= n; i++) {
    while (ch.size() >= 2) {
      long long blkg = ch.size();
      blkg--;
      if (solve(ch[blkg - 1], ch[blkg], isi[i]) <= 0) break;
      ch.pop_back();
    }
    ch.push_back(isi[i]);
  }
  n = ch.size();
  triangle maksi;
  for (int i = 0; i < n; i++) {
    long long kiri = i + 1;
    for (int j = i + 2; j < n; j++) {
      triangle cur = triangle(ch[i], ch[kiri], ch[j]);
      long long kanan = j - 1;
      for (; kiri < kanan; kiri++) {
        triangle nx = triangle(ch[i], ch[kiri + 1], ch[j]);
        if (nx < cur) break;
        cur = nx;
      }
      if (maksi < cur) maksi = cur;
    }
  }
  printans(maksi);
  return 0;
}