Codeforces Round #432 (Div. 2)

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 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：

在平面上有 a , b , c 三个点，问是否在平面上存在一点，使得a、b以该点为轴选择某一角度后a->b, b->c 

思路：

因为a可以到b，b可以到c，所以a可以到c且a,b,c三点在同一圆上，因为转的角度相同，所以ab = bc ，同时因为是在圆上，所以ac≠ab+bc

#include<iostream>#include<string.h>#include<stdio.h>#include<math.h>using namespace std;int main(){    double ax,bx,cx,ay,by,cy;    cin>>ax>>ay>>bx>>by>>cx>>cy;    double ab = sqrt(pow(ax-bx,2)+pow(ay-by,2));    double bc = sqrt(pow(cx-bx,2)+pow(cy-by,2));    double ac = sqrt(pow(ax-cx,2)+pow(ay-cy,2));    if(ab==bc && ac!=ab+bc) printf("Yes");    else printf("No");    return 0;}