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;
const int mo = 17;
int dv(long long int x) {
  x %= mo;
  return x < 0 ? x + mo : x;
}
int n, x[2022], y[2022];
long long int S;
int smo;
const int Z = 50000000;
vector<int> a[mo * mo];
vector<int> b[mo * mo][mo * mo];
bool test(int i, int j, int k) {
  long long int i1 = x[i], j1 = y[i];
  long long int i2 = x[j], j2 = y[j];
  long long int i3 = x[k], j3 = y[k];
  return i1 * j2 + i2 * j3 + i3 * j1 - j1 * i2 - j2 * i3 - j3 * i1 == S;
}
int main() {
  cin >> n >> S;
  S *= 2;
  smo = dv(S);
  for (int i = 0; i < n; i++) {
    scanf("%d%d", &x[i], &y[i]);
    a[dv(x[i]) * mo + dv(y[i])].push_back(i);
  }
  for (int i1 = 0; i1 < mo; i1++)
    for (int j1 = 0; j1 < mo; j1++)
      for (int i2 = 0; i2 < mo; i2++)
        for (int j2 = 0; j2 < mo; j2++)
          for (int i3 = 0; i3 < mo; i3++)
            for (int j3 = 0; j3 < mo; j3++)
              if (dv(i1 * j2 + i2 * j3 + i3 * j1 - j1 * i2 - j2 * i3 -
                     j3 * i1) == smo) {
                b[i1 * mo + j1][i2 * mo + j2].push_back(i3 * mo + j3);
              }
  for (int i = n - 1; i >= 0; i--) {
    int hi = dv(x[i]) * mo + dv(y[i]);
    a[hi].pop_back();
    int T = 0;
    for (int j = i - 1; j >= 0; j--) {
      int hj = dv(x[j]) * mo + dv(y[j]);
      for (vector<int>::iterator it = b[hi][hj].begin(); it != b[hi][hj].end();
           it++) {
        for (vector<int>::iterator it2 = a[*it].begin(); it2 != a[*it].end();
             it2++) {
          if (test(i, j, *it2)) {
            printf("Yes\n");
            printf("%d %d\n", x[i], y[i]);
            printf("%d %d\n", x[j], y[j]);
            printf("%d %d\n", x[*it2], y[*it2]);
            return 0;
          }
          ++T;
          if (T >= Z) break;
        }
        if (T >= Z) break;
      }
      if (T >= Z) break;
    }
  }
  puts("No");
  return 0;
}