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B. Arpa and an exam about geometry
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard 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.

注意三点不能连成一线

#include<cstdio>
int x[5],y[5];
int main(){
//  freopen("cfb.in","r",stdin);
    for (int i=1;i<=3;++i)scanf("%d%d",&x[i],&y[i]);
    long long tmp=(long long)(x[2]-x[1])*(x[2]-x[1])+(long long)(y[2]-y[1])*(y[2]-y[1]);
    long long tmp1=(long long)(x[2]-x[3])*(x[2]-x[3])+(long long)(y[2]-y[3])*(y[2]-y[3]);
    long long tmp2=(long long)(x[3]-x[1])*(x[3]-x[1])+(long long)(y[3]-y[1])*(y[3]-y[1]);
    double x1=(x[1]+x[3])*1.0/2;double y1=(y[1]+y[3])*1.0/2;
    if (x1==x[2]&&y1==y[2]) {printf("No");return 0;}
    if (tmp==tmp1) printf("Yes");else printf("No");
    return 0;
}