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 4*S*, 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.
In the first line of the input two integers *n* and *S* (3 ≤ *n* ≤ 5000, 1 ≤ *S* ≤ 10<sup class="upper-index">18</sup>) 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: *i*<sup class="upper-index">*th*</sup> of them consists of two integers *x*<sub class="lower-index">*i*</sub> and *y*<sub class="lower-index">*i*</sub> ( - 10<sup class="upper-index">8</sup> ≤ *x*<sub class="lower-index">*i*</sub>, *y*<sub class="lower-index">*i*</sub> ≤ 10<sup class="upper-index">8</sup>) — coordinates of *i*<sup class="upper-index">*th*</sup> point.

It is guaranteed that there is at least one triple of points not lying on the same line.
import heapq
import sys, os, io
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline

def f(u, v):
    return u * n + v

n, s = map(int, input().split())
xy = [tuple(map(int, input().split())) for _ in range(n)]
h = []
for i in range(n):
    xi, yi = xy[i]
    for j in range(i + 1, n):
        xj, yj = xy[j]
        d = (xi - xj) * (xi - xj) + (yi - yj) * (yi - yj)
        heapq.heappush(h, (d, f(i, j)))
        if len(h) > 2 * n:
            heapq.heappop()
ma = 0
z = []
for _, u in h:
    i, j = divmod(u, n)
    xi, yi = xy[i]
    xj, yj = xy[j]
    for k in range(n):
        xk, yk = xy[k]
        d = abs((xi - yk) * (yj - yk) - (xj - xk) * (yi - yk))
        if ma < d:
            ma, z = d, [i, j, k]
ans = []
for i in range(3):
    x1, y1 = xy[z[i]]
    x2, y2 = xy[z[(i + 1) % 3]]
    x3, y3 = xy[z[(i - 1) % 3]]
    x, y = x1 + x2 - x3, y1 + y2 - y3
    ans.append(" ".join(map(str, (x, y))))
sys.stdout.write("\n".join(ans))