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.io.InputStreamReader;
import java.io.BufferedReader;
import java.util.StringTokenizer;
public class BCF432A {
    public static void main(String[]args)throws Exception
    {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st=new StringTokenizer(br.readLine());
        long a1=Integer.parseInt(st.nextToken());
        long a2=Integer.parseInt(st.nextToken());
        long b1=Integer.parseInt(st.nextToken());
        long b2=Integer.parseInt(st.nextToken());
        long c1=Integer.parseInt(st.nextToken());
        long c2=Integer.parseInt(st.nextToken());
        long dab=dist(a1,a2,b1,b2);
        long dac=dist(a1,a2,c1,c2);
        long dbc=dist(b1,b2,c1,c2);
        if((b2-a2)*(c1-b1)==(c2-b2)*(b1-a1))
            System.out.println("No");
        else if(dab!=dbc)
            System.out.println("No");
        else
            System.out.println("Yes");
        
    }
    static Long dist(long a,long b,long c,long d)
    {
        long dd=((a-c)*(a-c))+((b-d)*(b-d));
        return dd;
    }
}