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_1
   nnasy */

#include <stdio.h>

//指定された構造体
typedef struct point{
    int x_elem;
    int y_elem;
}point;

//複素数平面上の2点a,bについて、bから見てaを270度回転させた先の点cを返す関数
point shift_point(point a, point b){
    point c;
    c.x_elem = b.x_elem + (a.y_elem - b.y_elem);
    c.y_elem = b.y_elem - (a.x_elem - b.x_elem);
    return c;
}

int main(void){
    //宣言
    point p1, p2, p3, p4;
    //入力
    scanf("%d %d %d %d", &p1.x_elem, &p1.y_elem, &p2.x_elem, &p2.y_elem);
    
    //座標(x3, y3)を求める
    p3 = shift_point(p1, p2);
    //座標(x4, y4)を求める
    p4 = shift_point(p2, p3);
    
    //出力
    printf("%d %d %d %d\n", p3.x_elem, p3.y_elem, p4.x_elem, p4.y_elem);
    
    return 0;
}