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.*;
import java.io.*;

public final class Main {

    public static void main(String[] args) throws Exception {
        PrintWriter pw = new PrintWriter(System.out);
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer(br.readLine());

        long ax = Long.parseLong(st.nextToken());
        long ay = Long.parseLong(st.nextToken());
        long bx = Long.parseLong(st.nextToken());
        long by = Long.parseLong(st.nextToken());
        long cx = Long.parseLong(st.nextToken());
        long cy = Long.parseLong(st.nextToken());

        if(ax == bx && bx == cx || ay == by && by == cy){
            System.out.println("No");
            return;
        }

        long distAToB = ((long)Math.pow(Math.abs(ax-bx) ,2) + (long)Math.pow(Math.abs(ay-by), 2));
        long distBToC = ((long)Math.pow(Math.abs(cx-bx) ,2) + (long)Math.pow(Math.abs(cy-by), 2));
        if( distAToB != distBToC){
            System.out.println("No");
            return;
        }
        //distance from a to b must be same as b to c but they cannot all lie on the same plane
        //ie the slope from a to b and b to c cannot be equal
        if((ax-bx) == 0 && (bx-cx) == 0){
            System.out.println("No");
            return;
        }else if((ax-bx) == 0 || (bx-cx) == 0){
            System.out.println("Yes");
            return;
        }
        double aToBSlope = (double)(ay-by)/(ax-bx);
        double bToCSlope = (double)(by-cy)/(bx-cx);

        pw.println(aToBSlope == bToCSlope ? "No" : "Yes");
        pw.close();
    }
}