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.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
import java.io.PrintWriter;
public class B851 
{
    static class Scanner
    {
        BufferedReader br;
        StringTokenizer tk=new StringTokenizer("");
        public Scanner(InputStream is) 
        {
            br=new BufferedReader(new InputStreamReader(is));
        }
        public int nextInt() throws IOException
        {
            if(tk.hasMoreTokens())
                return Integer.parseInt(tk.nextToken());
            tk=new StringTokenizer(br.readLine());
            return nextInt();
        }
        public long nextLong() throws IOException
        {
            if(tk.hasMoreTokens())
                return Long.parseLong(tk.nextToken());
            tk=new StringTokenizer(br.readLine());
            return nextLong();
        }
        public String next() throws IOException
        {
            if(tk.hasMoreTokens())
                return (tk.nextToken());
            tk=new StringTokenizer(br.readLine());
            return next();
        }
        public String nextLine() throws IOException
        {
            tk=new StringTokenizer("");
            return br.readLine();
        }
        public double nextDouble() throws IOException
        {
            if(tk.hasMoreTokens())
                return Double.parseDouble(tk.nextToken());
            tk=new StringTokenizer(br.readLine());
            return nextDouble();
        }
        public char nextChar() throws IOException
        {
            if(tk.hasMoreTokens())
                return (tk.nextToken().charAt(0));
            tk=new StringTokenizer(br.readLine());
            return nextChar();
        }
        public int[] nextIntArray(int n) throws IOException
        {
            int a[]=new int[n];
            for(int i=0;i<n;i++)
                a[i]=nextInt();
            return a;
        }
        public long[] nextLongArray(int n) throws IOException
        {
            long a[]=new long[n];
            for(int i=0;i<n;i++)
                a[i]=nextLong();
            return a;
        }
        public int[] nextIntArrayOneBased(int n) throws IOException
        {
            int a[]=new int[n+1];
            for(int i=1;i<=n;i++)
                a[i]=nextInt();
            return a;
        }
        public long[] nextLongArrayOneBased(int n) throws IOException
        {
            long a[]=new long[n+1];
            for(int i=1;i<=n;i++)
                a[i]=nextLong();
            return a;
        }
    
    
    }
    
    public static void main(String args[]) throws IOException
    {
        Scanner in=new Scanner(System.in);
        PrintWriter out=new PrintWriter(System.out);
        long ax=in.nextLong();
        long ay=in.nextLong();
        long bx=in.nextLong();
        long by=in.nextLong();
        long cx=in.nextLong();
        long cy=in.nextLong();
        if((bx-ax)*(bx-ax)+(by-ay)*(by-ay)==(cy-by)*(cy-by)+(cx-bx)*(cx-bx))
            if((bx-ax)*(cy-by)==(cx-bx)*(by-ay))
                    out.println("No");
            else
                    out.println("Yes");
        else
            out.println("No");
        out.close();
    }

}