Problem: Arpa is taking a geometry exam. Here is the last problem of the exam.

You are given three points a, b, c.

Find a point and an angle such that if we rotate the page around the point by the angle, the new position of a is the same as the old position of b, and the new position of b is the same as the old position of c.

Arpa is doubting if the problem has a solution or not (i.e. if there exists a point and an angle satisfying the condition). Help Arpa determine if the question has a solution or not.

Input

The only line contains six integers ax, ay, bx, by, cx, cy (|ax|, |ay|, |bx|, |by|, |cx|, |cy| ≤ 109). It's guaranteed that the points are distinct.

Output

Print "Yes" if the problem has a solution, "No" otherwise.

You can print each letter in any case (upper or lower).

Examples

Input

0 1 1 1 1 0


Output

Yes


Input

1 1 0 0 1000 1000


Output

No

Note

In the first sample test, rotate the page around (0.5, 0.5) by <image>.

In the second sample test, you can't find any solution.

Solution (in JAVA):

Solution:
//package c;

import java.math.BigDecimal;
import java.util.Scanner;

public class Practice {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner in = new Scanner(System.in);
		long xa = in.nextLong();
		long ya = in.nextLong();
		long xb = in.nextLong();
		long yb = in.nextLong();
		long xc = in.nextLong();
		long yc = in.nextLong();
		BigDecimal dis1 = new BigDecimal(xa-xb).pow(2).add(new BigDecimal(ya-yb).pow(2));
		BigDecimal dis2 = new BigDecimal(xb-xc).pow(2).add(new BigDecimal(yb-yc).pow(2));
		//double distance1 = Math.sqrt(Math.pow((xa-xb),2) + Math.pow((ya-yb), 2));
		//double distance2 = Math.sqrt(Math.pow((xc-xb),2) + Math.pow((yc-yb), 2));
		if(!dis1.equals(dis2)) System.out.println("No");
		else {
			if((xa-xb)*(yb-yc) != (xb-xc)*(ya-yb))
				System.out.println("Yes");
			else System.out.println("No");
		}
		in.close();
	}

}