Codeforces Round #432 (Div. 2, based on IndiaHacks Final Round 2017) B

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 .

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

题意：三个点，从A B C找到一个圆形，通过旋转可以使得A到B这个位置，B到C这个位置

解法：

1 AB BC距离相等

2 大概想到是一个.....扇子..不能符合条件的只有三点共线的情况

 #include<bits/stdc++.h>
 using namespace std;
 struct Node{
     long long x,y;
 }node[];
 long long dis(Node a,Node b){
     return (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y);
 }
 int main(){
     for(int i=;i<=;i++){
         cin>>node[i].x>>node[i].y;
     }

     long long a=(node[].x-node[].x)*(node[].y-node[].y);
     long long b=(node[].y-node[].y)*(node[].x-node[].x);
     if(a!=b&&(dis(node[],node[])==dis(node[],node[]))){
         cout<<"Yes"<<endl;
     }else{
         cout<<"No"<<endl;
     }
     return ;
 }