Problem: There is a strange peculiarity: if you connect the cities of Rostov, Taganrog and Shakhty, peculiarly, you get a triangle

«Unbelievable But True»

Students from many different parts of Russia and abroad come to Summer Informatics School. You marked the hometowns of the SIS participants on a map.

Now you decided to prepare an interesting infographic based on this map. The first thing you chose to do is to find three cities on this map, such that they form a triangle with area S.

Input

The first line of input contains two integers n and S (3 ≤ n ≤ 2000, 1 ≤ S ≤ 2 ⋅ 10^{18}) — the number of cities on the map and the area of the triangle to be found.

The next n lines contain descriptions of the cities, one per line. Each city is described by its integer coordinates x_i, y_i (-10^9 ≤ x_i, y_i ≤ 10^9). 

It is guaranteed that all cities are located at distinct points. It is also guaranteed that no three cities lie on the same line.

Output

If the solution doesn't exist — print «No».

Otherwise, print «Yes», followed by three pairs of coordinates (x, y) — the locations of the three cities, which form the triangle of area S.

Examples

Input

3 7
0 0
3 0
0 4


Output

No


Input

4 3
0 0
2 0
1 2
1 3


Output

Yes
0 0
1 3
2 0

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int n, Vsize;
long long S;
long long x[2007], y[2007];
pair<long long, long long> V[4000007];
int P[2007];
int where[2007];
int nv;
inline int delta(pair<long long, long long> p) {
  if (p.first > 0 && p.second >= 0) return 1;
  if (p.first <= 0 && p.second > 0) return 2;
  if (p.first < 0 && p.second <= 0) return 3;
  return 4;
}
inline pair<long long, long long> normal(pair<long long, long long> v) {
  return {y[v.first] - y[v.second], x[v.second] - x[v.first]};
}
inline pair<long long, long long> point(int p) { return {x[p], y[p]}; }
constexpr long long cross(pair<long long, long long> a,
                          pair<long long, long long> b) {
  return a.first * b.second - a.second * b.first;
}
constexpr long long dot(pair<long long, long long> a,
                        pair<long long, long long> b) {
  return a.first * b.first + a.second * b.second;
}
bool angle_comp(const pair<long long, long long> &a,
                const pair<long long, long long> &b) {
  if (delta(a) != delta(b)) return delta(a) < delta(b);
  return cross(a, b) > 0;
}
bool comp(const pair<long long, long long> &a,
          const pair<long long, long long> &b) {
  return angle_comp(normal(a), normal(b));
}
bool comp2(const int a, const int b) {
  return dot(point(a), normal(V[nv])) < dot(point(b), normal(V[nv]));
}
int main() {
  scanf("%d %I64d", &n, &S);
  for (int i = 1; i <= n; i++) scanf("%I64d %I64d", &x[i], &y[i]);
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      if (i != j) {
        V[Vsize++] = {i, j};
      }
    }
  }
  sort(V, V + Vsize, comp);
  for (int i = 1; i <= n; i++) P[i - 1] = i;
  nv = 0;
  sort(P, P + n, comp2);
  for (int i = 0; i < n; i++) where[P[i]] = i;
  for (int i = 0; i < Vsize; i++) {
    long long x = ::x[V[i].first];
    long long y = ::y[V[i].first];
    long long x1 = ::x[V[i].second];
    long long y1 = ::y[V[i].second];
    long long ldot = 2LL * S + y * (x1 - x) - x * (y1 - y);
    int pocz = 0, kon = n - 1, mid;
    bool found = false;
    while (pocz <= kon) {
      mid = (pocz + kon) / 2;
      long long d = dot(point(P[mid]), normal(V[i]));
      if (d == ldot) {
        found = true;
        break;
      }
      if (d < ldot)
        pocz = mid + 1;
      else
        kon = mid - 1;
    }
    if (found) {
      printf("Yes\n");
      printf("%I64d %I64d\n%I64d %I64d\n%I64d %I64d\n", x, y, x1, y1,
             ::x[P[mid]], ::y[P[mid]]);
      return 0;
    }
    int a = V[i].first;
    int b = V[i].second;
    if (i == 0) {
      if ((where[a] < where[b] &&
           dot(point(a), normal(V[i + 1])) > dot(point(b), normal(V[i + 1]))) ||
          (where[a] > where[b] &&
           dot(point(a), normal(V[i + 1])) < dot(point(b), normal(V[i + 1])))) {
        swap(P[where[a]], P[where[b]]);
        swap(where[a], where[b]);
      }
    } else {
      swap(P[where[a]], P[where[b]]);
      swap(where[a], where[b]);
    }
  }
  printf("No\n");
  return 0;
}