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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;

#define inf 1072114514
#define llinf 4154118101919364364
#define mod 1000000007
#define pi 3.1415926535897932384


int round(int a,int b){if((a%b)*2 >= b){return (a/b)+1;}return a/b;} 
int gcd(int a,int b){int c;while(b!=0){c=a%b;a=b;b=c;}return a;}    //最大公約数
int lcm(int a,int b){int c=gcd(a,b);a/=c;return a*b;}     //最小公倍数
int nCr(int a,int b){int i,r=1;for(i=1;i<=b;i++){r*=(a+1-i);r/=i;}return r;} //コンビネーション
int nHr(int a,int b){return nCr(a+b-1,b);}       // 重複組み合わせ
int fact(int a){int i,r=1;for(i=1;i<=a;i++){r*=i;}return r;}  //階乗
int pow(int a,int b){int i,r=1;for(i=1;i<=b;i++){r*=a;}return r;} // a~bまでの階乗
int dsum(int x){int r=0;while(x){r+=(x%10);x/=10;}return r;}   //数字の各位の和
int dsumb(int x,int b){int r=0;while(x){r+=(x%b);x/=b;}return r;}  // b進数の各位の和？
int sankaku(int x){return ((1+x)*x)/2;} //三角数 xまでの和

//以下long long ver
long long llmax(long long a,long long b){if(a>b){return a;}return b;}
long long llmin(long long a,long long b){if(a<b){return a;}return b;}
long long llzt(long long a,long long b){return llmax(a,b)-llmin(a,b);}
long long llround(long long a,long long b){if((a%b)*2 >= b){return (a/b)+1;}return a/b;}
long long llceil(long long a,long long b){if(a%b==0){return a/b;}return (a/b)+1;}
long long llgcd(long long a,long long b){long long c;while(b!=0){c=a%b;a=b;b=c;}return a;}
long long lllcm(long long a,long long b){long long c=llgcd(a,b);a/=c;return a*b;}
long long llnCr(long long a,long long b){long long i,r=1;for(i=1;i<=b;i++){r*=(a+1-i);r/=i;}return r;}
long long llnHr(long long a,long long b){return llnCr(a+b-1,b);}
long long llfact(long long a){long long i,r=1;for(i=1;i<=a;i++){r*=i;}return r;}
long long llpow(long long a,long long b){long long i,r=1;for(i=1;i<=b;i++){r*=a;}return r;}
long long lldsum(long long x){long long r=0;while(x){r+=(x%10);x/=10;}return r;}
long long lldsumb(long long x,long long b){long long r=0;while(x){r+=(x%b);x/=b;}return r;}
long long llsankaku(long long x){return ((1+x)*x)/2;}

//double
double dbmax(double a,double b){if(a>b){return a;}return b;}
double dbmin(double a,double b){if(a<b){return a;}return b;}
double dbzt(double a,double b){return dbmax(a,b)-dbmin(a,b);}


typedef pair<int, int> ii;
typedef long long ll;
typedef vector<ll> vll;
typedef pair<ll,ll> pll;
#define forr(i,a,b) ; for(int i=(a); i<(b); i++)
#define clean(arr,val) memset(arr,val,sizeof(arr))
#define forn(i,n) forr(i,0,n)
#define PB push_back
typedef vector<int> vi;
typedef vector<ii> vii;
typedef vector<pll> vpll;
const int dx[8] = {1, 0, -1, 0, 1, -1, -1, 1};
const int dy[8] = {0, 1, 0, -1, 1, 1, -1, -1};
bool is_prime (int x){
  if (x<=1) return false ;
  for (int i=2 ; i*i<=x ; i++){
    if (x%i==0) return false ;
  }
   return true;
}

 
/*CODE START HERE*/
int main() {
  int x_1 , y_1 , x_2 , y_2, x_3 , y_3 , x_4 , y_4;
  cin >> x_1 >> y_1 >> x_2 >> y_2;
  int l ;
  if ( x_1 == x_2 ){
    l = abs(y_2 - y_1) ;
    if ( y_2 > y_1 ) {
      x_4 = -l+x_1;
      x_4 = x_3 ;
      y_3 = y_2;
      y_4 = y_1;
    }else  if ( y_2 < y_1 ) {
      x_3 = l+x_2;
      x_4 = x_3 ;
      y_3 = y_2;
      y_4 = y_1;
    }
  }else if ( y_1 == y_2 ){
    l = abs(x_2 - x_1) ;
    if ( x_2 > x_1 ) {
      y_3 = l+y_2;
      y_4 = y_3 ;
      x_3 = x_2;
      x_4 = x_1;
    }else  if ( x_2 < x_1 ) {
      y_3 = -l+y_2;
      y_4 = y_3 ;
      x_3 = x_2;
      x_4 = x_1;
    }
  }
  cout << x_3 <<  ' ' << y_3 << ' ' << x_4 << ' ' << y_4 << endl;
}