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 Python (3.4.3) (judge verdict: Wrong Answer)

import math
x1,y1,x2,y2=map(int,input().split())
#dist=sqrt((x2-x1)**2+(y2-y1)**2)
if x1==x2:
    dist=abs(y2-y1)
    print('{} {} {} {}'.format(x2-dist,y2,x1-dist,y1))
elif y1==y2:
    dist=abs(x2-x1)
    print('{} {} {} {}'.format(x2,y2-dist,x1,y1-dist))
else:
    x3=x2-y2+y1
    y3=x2-x1+y2
    x4=x1+y1-y2
    y4=x2+y1-x1
    #x3=x1+((x2-x1)*(math.sqrt(2)/2)-(y2-y1)*(math.sqrt(2)/2))*math.sqrt(2)
    #y3=y1+((x2-x1)*(math.sqrt(2)/2)-(y2-y1)*(math.sqrt(2)/2))*math.sqrt(2)
    #x4=x2+((x1-x2)*(math.sqrt(2)/2)-(y1-y2)*(math.sqrt(2)/2))*math.sqrt(2)
    #y4=y2+((x1-x2)*(math.sqrt(2)/2)-(y1-y2)*(math.sqrt(2)/2))*math.sqrt(2)
    print('{} {} {} {}'.format(x3,y3,x4,y4))