Pashmak has fallen in love with an attractive girl called Parmida since one year ago...

Today, Pashmak set up a meeting with his partner in a romantic garden. Unfortunately, Pashmak has forgotten where the garden is. But he remembers that the garden looks like a square with sides parallel to the coordinate axes. He also remembers that there is exactly one tree on each vertex of the square. Now, Pashmak knows the position of only two of the trees. Help him to find the position of two remaining ones.

Input

The first line contains four space-separated x1, y1, x2, y2 ( - 100 ≤ x1, y1, x2, y2 ≤ 100) integers, where x1 and y1 are coordinates of the first tree and x2 and y2 are coordinates of the second tree. It's guaranteed that the given points are distinct.

Output

If there is no solution to the problem, print -1. Otherwise print four space-separated integers x3, y3, x4, y4 that correspond to the coordinates of the two other trees. If there are several solutions you can output any of them. 

Note that x3, y3, x4, y4 must be in the range ( - 1000 ≤ x3, y3, x4, y4 ≤ 1000).

Examples

Input

0 0 0 1


Output

1 0 1 1


Input

0 0 1 1


Output

0 1 1 0


Input

0 0 1 2


Output

-1

//Omar brome, lcpc practice, codeforces: *
//basic code for fast java input/output learned from kattis

import java.util.*;
import java.io.*;

public class Main{
public static void main(String[] args){
Kattio tuna= new Kattio(System.in,System.out);
int x1=tuna.getInt();
int y1=tuna.getInt();

int x2=tuna.getInt();
int y2=tuna.getInt();

if(Math.abs(x2-x1)==Math.abs(y2-y1))
    tuna.println(x1+" "+y2+" "+x2+" "+y1);
else if (x1==x2)
    tuna.println((x1+y2-y1)+" "+y1+" "+(x1+y2-y1)+" "+y2);
else if (y1==y2)
    tuna.println(x1+" "+(y1+x2-x1)+" "+x2+" "+(y1+x2-x1));
else
    tuna.println("-1");
tuna.flush();
}
}

class Kattio extends PrintWriter {
  public Kattio(InputStream i) {
    super(new BufferedOutputStream(System.out));
    r = new BufferedReader(new InputStreamReader(i));
  }
  public Kattio(InputStream i, OutputStream o) {
    super(new BufferedOutputStream(o));
    r = new BufferedReader(new InputStreamReader(i));
  }

  public boolean hasMoreTokens() {
    return peekToken() != null;
  }

  public int getInt() {
    return Integer.parseInt(nextToken());
  }

  public double getDouble() { 
    return Double.parseDouble(nextToken());
  }

  public long getLong() {
    return Long.parseLong(nextToken());
  }

  public String getWord() {
    return nextToken();
  }



  private BufferedReader r;
  private String line;
  private StringTokenizer st;
  private String token;

  private String peekToken() {
    if (token == null) 
        try {
        while (st == null || !st.hasMoreTokens()) {
            line = r.readLine();
            if (line == null) return null;
            st = new StringTokenizer(line);
        }
        token = st.nextToken();
        } catch (IOException e) { }
    return token;
  }

  private String nextToken() {
    String ans = peekToken();
    token = null;
    return ans;
  }
}
