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 9.2.1) (judge verdict: Accepted)

/*	ex7-1
	toshiwaka	*/

typedef struct point{
    int x_elem;
    int y_elem;
}point;

#include <stdio.h>

int main(void) {
  //point:正方形の頂点
  //diff:2点間のx,y軸方向の距離
  point point[4], diff;

  //入力
  scanf("%d %d %d %d", &point[0].x_elem, &point[0].y_elem, &point[1].x_elem, &point[1].y_elem);

  //2点間のx,y軸方向の距離
  diff.x_elem = point[1].x_elem - point[0].x_elem;
  diff.y_elem = point[1].y_elem - point[0].y_elem;

  //残りの２点を算出
  point[2].x_elem = point[1].x_elem - diff.y_elem;
  point[2].y_elem = point[1].y_elem + diff.x_elem;
  point[3].x_elem = point[0].x_elem - diff.y_elem;
  point[3].y_elem = point[0].y_elem + diff.x_elem;

  //出力
  printf("%d %d %d %d\n", point[2].x_elem, point[2].y_elem, point[3].x_elem, point[3].y_elem);

  return 0;
}