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;
const long long MOD = 1e9 + 7;
const long long N = 5e3 + 5;
long long power(long long a, long long b) {
  long long ret = 1;
  while (b) {
    if (b & 1) ret *= a;
    a *= a;
    if (ret >= MOD) ret %= MOD;
    if (a >= MOD) a %= MOD;
    b >>= 1;
  }
  return ret;
}
long long invmod(long long x) { return power(x, MOD - 2); }
int32_t main() {
  long long a[4], b[4];
  for (long long i = 0; i < 3; i++) {
    cin >> a[i] >> b[i];
  }
  double x = (double)sqrt((a[1] - a[0]) * (a[1] - a[0]) +
                          (b[1] - b[0]) * (b[1] - b[0]));
  double y = (double)sqrt((a[1] - a[2]) * (a[1] - a[2]) +
                          (b[1] - b[2]) * (b[1] - b[2]));
  double z = (double)sqrt((a[2] - a[0]) * (a[2] - a[0]) +
                          (b[2] - b[0]) * (b[2] - b[0]));
  if (x + y > z && y + z > x && x + z > y && x == y) {
    cout << "Yes" << endl;
  } else {
    cout << "No" << endl;
  }
}