CodeForces - 851B -Arpa and an exam about geometry（计算几何）

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.

题解：若三个点可以通过旋转使a到b，b到c，那只需要两个条件，共圆和弧长相等，共圆可以转换为AB和BC不在一条直线上，即AB，BC斜率不相等，弧长相等转换为弦相等，及AB之间的距离等于BC之间的距离

代码：

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>

using namespace std;

int main() {

	long  double ax,ay,bx,by,cx,cy;
	cin>>ax>>ay>>bx>>by>>cx>>cy;
	if((by-ay)/(bx-ax)!=(cy-by)/(cx-bx)&&(by-ay)*(by-ay)+(bx-ax)*(bx-ax)==(cy-by)*(cy-by)+(cx-bx)*(cx-bx)) {
		printf("Yes\n");
	} else {
		printf("No\n");
	}

	return 0;
}