Problem:
Score : 200 points

Problem Statement

There is a square in the xy-plane. The coordinates of its four vertices are (x_1,y_1),(x_2,y_2),(x_3,y_3) and (x_4,y_4) in counter-clockwise order.
(Assume that the positive x-axis points right, and the positive y-axis points up.)

Takahashi remembers (x_1,y_1) and (x_2,y_2), but he has forgot (x_3,y_3) and (x_4,y_4).

Given x_1,x_2,y_1,y_2, restore x_3,y_3,x_4,y_4. It can be shown that x_3,y_3,x_4 and y_4 uniquely exist and have integer values.

Constraints

|x_1|,|y_1|,|x_2|,|y_2| \leq 100

(x_1,y_1) ≠ (x_2,y_2)

All values in input are integers.

Input

Input is given from Standard Input in the following format:

x_1 y_1 x_2 y_2

Output

Print x_3,y_3,x_4 and y_4 as integers, in this order.

Sample Input 1

0 0 0 1

Sample Output 1

-1 1 -1 0

(0,0),(0,1),(-1,1),(-1,0) is the four vertices of a square in counter-clockwise order.
Note that (x_3,y_3)=(1,1),(x_4,y_4)=(1,0) is not accepted, as the vertices are in clockwise order.

Sample Input 2

2 3 6 6

Sample Output 2

3 10 -1 7

Sample Input 3

31 -41 -59 26

Sample Output 3

-126 -64 -36 -131

Solution in Ruby (2.3.3) (judge verdict: Accepted)

x1, y1, x2, y2 = gets.split.map(&:to_i)

dx = (x2 - x1)
dy = (y2 - y1)

if dx < 0
  dx *= -1
end

if dy < 0
  dy *= -1
end

if x1 < x2 && y1 <= y2
  puts "#{x2-dy} #{y2+dx} #{x1-dy} #{y1+dx}"
elsif x1 >= x2 && y1 < y2
  puts "#{x2-dy} #{y2-dx} #{x1-dy} #{y1-dx}"
elsif x1 > x2 && y1 >= y2
  puts "#{x2+dy} #{y2-dx} #{x1+dy} #{y1-dx}"
else
  puts "#{x2+dy} #{y2+dx} #{x1+dy} #{y1+dx}"
end