B. Arpa and an exam about geometry

B. Arpa and an exam about geometry

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 a_x, a_y, b_x, b_y, c_x, c_y (|a_x|, |a_y|, |b_x|, |b_y|, |c_x|, |c_y| ≤ 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 .

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

 

#include <bits/stdc++.h>
using namespace std;
int main(){
  long long int ax,ay,bx,by,cx,cy;
  cin>>ax>>ay>>bx>>by>>cx>>cy;
  ax*=2;
  ay*=2;
  bx*=2;
  by*=2;
  cx*=2;
  cy*=2;
  long long int n,m,i,j;
  n=abs(ax-bx)*abs(ax-bx)+abs(ay-by)*abs(ay-by);
  m=abs(bx-cx)*abs(bx-cx)+abs(by-cy)*abs(by-cy);
  i=(ax+cx)/2;
  j=(ay+cy)/2;
  if(n==m&&!(i==bx&&j==by)){
    cout<<"Yes"<<endl;
  }else{
    cout<<"No"<<endl;
  }
return 0;
}