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:
import java.util.Arrays;
import java.util.HashSet;
import java.util.Scanner;
import java.util.Set;

public class Main {

    static long distance(int x, int y, int xx, int yy) {
        return (long) (x - xx) * (x - xx) + (y - yy) * (y - yy);
    }

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int ax = in.nextInt(), ay = in.nextInt();
        int bx = in.nextInt(), by = in.nextInt();
        int cx = in.nextInt(), cy = in.nextInt();
        if ((long) (ax - bx) == (bx - cx) && (long) (ay - by) == (by - cy)) {
            System.out.println("No");
        } else if ((long) (ax-bx)*(ax-bx) + (long) (ay-by)*(ay-by) == (long) (bx-cx)*(bx-cx) + (long) (by-cy)*(by-cy)){
            System.out.println("Yes");
        } else {
            System.out.println("No");
        }
    }
}