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;
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC optimize("unroll-loops")
const int MOD = (int)1e9 + 7;
const int FFTMOD = 1007681537;
const int INF = (int)1e9;
const long long LINF = (long long)1e18;
const long double PI = acos((long double)-1);
const long double EPS = 1e-9;
inline long long gcd(long long a, long long b) {
  long long r;
  while (b) {
    r = a % b;
    a = b;
    b = r;
  }
  return a;
}
inline long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
inline long long fpow(long long n, long long k, int p = MOD) {
  long long r = 1;
  for (; k; k >>= 1) {
    if (k & 1) r = r * n % p;
    n = n * n % p;
  }
  return r;
}
template <class T>
inline int chkmin(T& a, const T& val) {
  return val < a ? a = val, 1 : 0;
}
template <class T>
inline int chkmax(T& a, const T& val) {
  return a < val ? a = val, 1 : 0;
}
inline long long isqrt(long long k) {
  long long r = sqrt(k) + 1;
  while (r * r > k) r--;
  return r;
}
inline long long icbrt(long long k) {
  long long r = cbrt(k) + 1;
  while (r * r * r > k) r--;
  return r;
}
inline void addmod(int& a, int val, int p = MOD) {
  if ((a = (a + val)) >= p) a -= p;
}
inline void submod(int& a, int val, int p = MOD) {
  if ((a = (a - val)) < 0) a += p;
}
inline int mult(int a, int b, int p = MOD) { return (long long)a * b % p; }
inline int inv(int a, int p = MOD) { return fpow(a, p - 2, p); }
inline int sign(long double x) { return x < -EPS ? -1 : x > +EPS; }
inline int sign(long double x, long double y) { return sign(x - y); }
const int maxn = 2000 + 5;
int n;
long long S;
int x[maxn];
int y[maxn];
long long f[maxn][maxn];
void volamtruyenkyii() {
  cin >> n >> S, S = S * 2;
  for (int i = (0); i < (n); ++i) cin >> x[i] >> y[i];
  for (int i = (0); i < (n); ++i)
    for (int j = (0); j < (n); ++j)
      f[i][j] = (long long)x[i] * y[j] - (long long)x[j] * y[i];
  for (int i = (0); i < (n); ++i) {
    for (int j = (i + 1); j < (n); ++j) {
      int ix = -1;
      long long t1 = S - f[i][j];
      long long t2 = -S - f[i][j];
      int ptr = j + 1;
      while (ptr < n % 4) {
        long long x0 = f[j][ptr] - f[i][ptr];
        ix = x0 == t1 || x0 == t2 ? ptr : ix;
        ptr++;
      }
      for (int k = ptr; k < n; k += 4) {
        long long x0 = f[j][k] - f[i][k];
        ix = x0 == t1 || x0 == t2 ? k : ix;
        long long x1 = f[j][k + 1] - f[i][k + 1];
        ix = x1 == t1 || x1 == t2 ? k + 1 : ix;
        long long x2 = f[j][k + 2] - f[i][k + 2];
        ix = x2 == t1 || x2 == t2 ? k + 2 : ix;
        long long x3 = f[j][k + 3] - f[i][k + 3];
        ix = x3 == t1 || x3 == t2 ? k + 3 : ix;
      }
      if (ix != -1) {
        cout << "Yes\n";
        cout << x[i] << " " << y[i] << "\n";
        cout << x[j] << " " << y[j] << "\n";
        cout << x[ix] << " " << y[ix] << "\n";
        return;
      }
    }
  }
  cout << "No\n";
}
int main(int argc, char* argv[]) {
  ios_base::sync_with_stdio(0), cin.tie(0);
  if (argc > 1) {
    assert(freopen(argv[1], "r", stdin));
  }
  if (argc > 2) {
    assert(freopen(argv[2], "wb", stdout));
  }
  volamtruyenkyii();
  cerr << "\nTime elapsed: " << 1000 * clock() / CLOCKS_PER_SEC << "ms\n";
  return 0;
}