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 C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>

int main(int argc, char const *argv[])
{
	int x[8], vec[8];
	scanf("%d%d%d%d", &x[0], &x[1], &x[2], &x[3]);
	vec[0]=x[2]-x[0];
	vec[1]=x[3]-x[1];
	for (int i = 2; i <= 6; i+=2)
	{
		vec[i]=-vec[i-1];
		vec[i+1]=vec[i-2];
	}
	for (int i = 4; i <=6 ; i+=2)
	{
		x[i]=x[i-1]+vec[i-2];
		x[i+1]=x[i-2]+vec[i-1];
	}
	printf("%d %d %d %d", x[4], x[5], x[6], x[7]);
	return 0;
}