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)

#include <stdio.h>

typedef struct point{ // 点を表す構造体
    int x_elem;
    int y_elem;
}point;

point shift_point(point vtx1, point vtx2){ // vtx1を中心にvtx2を90度回転させた座標を返す関数
    /* 複素数平面上で考える。
     * 複素数 z1 を中心に複素数 z2 を90°回転させた複素数 z は
     *   z = (z2 - z1)i + z1  と表される。
     *
     * z1 = x1 + iy1, z2 = x2 + iy2 とすると
     * z = {x2 + iy2 - (x1 + iy1)}i + x1 + iy1
     *   = {x2 - x1 + i(y2 - y1)}i + x1 + iy1
     *   = -(y2 - y1) + i(x2 - x1) + x1 + iy1
     *   = x1 + y1 - y2 + i(- x1 + x2 + y1)
     */
    point vtx;
    vtx.x_elem = vtx1.x_elem + vtx1.y_elem - vtx2.y_elem;
    vtx.y_elem = - vtx1.x_elem + vtx2.x_elem + vtx1.y_elem;
    return vtx;
}

int main(void){
    point vtx1, vtx2, vtx3, vtx4;
    scanf("%d%d%d%d", &vtx1.x_elem, &vtx1.y_elem, &vtx2.x_elem, &vtx2.y_elem);
    
    // vtx1, vtx2, vtx3, vtx4 は反時計回りに並んでいる
    vtx4 = shift_point(vtx1, vtx2); // vtx4はvtx1を中心にvtx2を90°回転させた座標にある。
    vtx3 = shift_point(vtx4, vtx1); // vtx3はvtx4を中心にvtx1を90°回転させた座標にある。
  
    printf("%d %d %d %d\n", vtx3.x_elem, vtx3.y_elem, vtx4.x_elem, vtx4.y_elem);
    return 0;
}