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>
using namespace std;
int main() {
  long long int ax, ay, bx, by, cx, cy;
  scanf("%lld%lld%lld%lld%lld%lld", &ax, &ay, &bx, &by, &cx, &cy);
  long long int l1 = ((by - ay) * (by - ay) + (bx - ax) * (bx - ax));
  long long int l2 = ((cy - by) * (cy - by) + (cx - bx) * (cx - bx));
  int mark = 0;
  if ((ax == bx && ax == cx) || (ay == by && ay == cy))
    mark = 1;
  else {
    double k1 = (by - ay) * 1.0 / (bx - ax);
    double k2 = (cy - by) * 1.0 / (cx - bx);
    if (k1 == k2) mark = 1;
  }
  if (!mark && l1 == l2)
    printf("Yes\n");
  else
    printf("No\n");
  return 0;
}