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)

/* ex 7_2
   offNaria */

#include <stdio.h>

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

//b-aベクトルをaを中心として反時計まわりに90度回転し，始点をbにしたものの終点を計算する関数
point shift_point(point a, point b);

int main(void)
{
    point p_1, p_2, p_3, p_4;
    scanf("%d %d %d %d", &p_1.x_elem, &p_1.y_elem, &p_2.x_elem, &p_2.y_elem);
    p_3 = shift_point(p_1, p_2); //p_2-p_1ベクトルをp_1を中心として反時計回りに90度回転し，始点をp_2としたベクトルの終点がp_3
    p_4 = shift_point(p_2, p_3); //p_3-p_2ベクトルをp_2を中心として反時計回りに90度回転し，始点をp_3としたベクトルの終点がp_4
    printf("%d %d %d %d\n", p_3.x_elem, p_3.y_elem, p_4.x_elem, p_4.y_elem);
    return 0;
}

point shift_point(point a, point b)
{
    point rotatedVec = {b.x_elem + (a.y_elem - b.y_elem), b.y_elem + (b.x_elem - a.x_elem)};
    return rotatedVec;
}