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;
long double a[3][2];
long double dis(int x, int y) {
  long double ans = 0;
  for (int i = 0; i < 2; i++) {
    if (a[x][i] > a[y][i])
      ans += ((a[x][i] - a[y][i]) * (a[x][i] - a[y][i]));
    else
      ans += ((a[y][i] - a[x][i]) * (a[y][i] - a[x][i]));
  }
  return ans;
}
bool iscollinear() {
  return (((a[2][0] - a[1][0]) * (a[1][1] - a[0][1])) ==
          ((a[2][1] - a[1][1]) * (a[1][0] - a[0][0])));
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  for (int i = 0; i < 3; i++) {
    for (int j = 0; j < 2; j++) cin >> a[i][j];
  }
  if (dis(0, 1) == dis(1, 2) && !iscollinear())
    cout << "yes" << endl;
  else
    cout << "no" << endl;
  return 0;
}