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;
double dx, dy;
long double x[10], y[10];
long double dist(int i, int j) {
  return sqrtl((x[i] - x[j]) * (x[i] - x[j]) + (y[i] - y[j]) * (y[i] - y[j]));
}
int main() {
  double EPS = std::numeric_limits<long double>::epsilon();
  for (int i = 0; i < 3; i++) {
    scanf(" %lf %lf", &dx, &dy);
    x[i] = dx;
    y[i] = dy;
  }
  long double a = dist(0, 1), b = dist(1, 2), c = dist(0, 2);
  long double mx = a;
  if (fabs(a - b) < EPS && fabs(c - (a + b)) > EPS) {
    printf("YES\n");
  } else {
    printf("NO\n");
  }
  return 0;
}