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
    Jun_JPN */

#include <stdio.h>

typedef struct{ // 頂点の(x,y)座標
    int x_elem;
    int y_elem;
}point;

point shift_point(point A_vec, point B_vec, int pm);    // pm = 1 or -1

int main(void){

    point A_vec, B_vec, C_vec, D_vec;
    scanf("%d %d %d %d", &A_vec.x_elem, &A_vec.y_elem, &B_vec.x_elem, &B_vec.y_elem);

    C_vec = shift_point(B_vec, A_vec, -1);  // BAベクトルを-90度回転
    D_vec = shift_point(A_vec, B_vec,  1);  // ABベクトルを+90度回転

    printf("%d %d %d %d\n", C_vec.x_elem, C_vec.y_elem, D_vec.x_elem, D_vec.y_elem);

    return 0;
}

/* ABベクトルをpm*90度回転した先にある点の座標を返す関数 */
point shift_point(point A_vec, point B_vec, int pm){
    point AB_vec, AD_vec, D_vec;

    /* 引数OAベクトル，OBベクトルからABベクトルを計算 */
    AB_vec.x_elem = B_vec.x_elem - A_vec.x_elem;
    AB_vec.y_elem = B_vec.y_elem - A_vec.y_elem;

    /* ABベクトルを90度回転し，ADベクトルを得る
        ただしpm=-1のときは-90度回転となる */
    AD_vec.x_elem = (-1)*pm*AB_vec.y_elem;
    AD_vec.y_elem = pm*AB_vec.x_elem;

    /* OAベクトルとADベクトルの和を取ればODベクトルが得られる */
    D_vec.x_elem = A_vec.x_elem + AD_vec.x_elem;
    D_vec.y_elem = A_vec.y_elem + AD_vec.y_elem;

    return D_vec;
}