Problem: 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 <image>.

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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
int main() {
  long ax, ay, bx, by, cx, cy;
  double dx, dy;
  bool vis1, vis2;
  scanf("%ld%ld%ld%ld%ld%ld", &ax, &ay, &bx, &by, &cx, &cy);
  if (ax == cx && ay == cy) {
    printf("Yes\n");
    return 0;
  }
  if (ax + cx == bx * 2 && ay + cy == by * 2) {
    printf("No\n");
    return 0;
  }
  if (cx - ax != 0)
    vis1 = true;
  else
    vis1 = false;
  dx = 1.0 * (ax + cx) / 2;
  dy = 1.0 * (ay + cy) / 2;
  if (fabs(bx - dx) > 0.00000000001)
    vis2 = true;
  else
    vis2 = false;
  if ((vis1 == false && vis2 == true && fabs(by - dy) < 0.00000000001) ||
      (vis1 == true && fabs(cy - ay) < 0.00000000001 && vis2 == false) ||
      (vis1 == true && vis2 == true &&
       (cy - ay) * (by - dy) == -(cx - ax) * (bx - dx)))
    printf("Yes\n");
  else
    printf("No\n");
  return 0;
}