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
//TKC001

#include<stdio.h>
//pointの宣言
typedef struct point
{
    int x_elem;
    int y_elem;
}point;

//点Pをx軸方向にx_diff, y軸方向にy_diffだけずらす
point shift_point(point p, int x_diff, int y_diff){
    p.x_elem += x_diff;
    p.y_elem += y_diff;

    return p;
}


int main(void){
    point poi1, poi2, poi3, poi4;
    int x_diff, y_diff;
    //入力
    scanf("%d %d %d %d", &poi1.x_elem, &poi1.y_elem, &poi2.x_elem, &poi2.y_elem);

    //(x1, y1)と(x2, y2)はx軸方向にx_diff, y軸方向にy_diffだけずらせばそれぞれ(x3, y3), (x4, y4)になる。
    x_diff = poi1.y_elem-poi2.y_elem;
    y_diff = poi2.x_elem-poi1.x_elem;

    //ずらす処理
    poi3 = shift_point(poi2, x_diff, y_diff);
    poi4 = shift_point(poi1, x_diff, y_diff);

    //出力
    printf("%d %d %d %d", poi3.x_elem, poi3.y_elem, poi4.x_elem, poi4.y_elem);

    return 0;
}