At a geometry lesson Gerald was given a task: to get vector B out of vector A. Besides, the teacher permitted him to perform the following operations with vector А:

  * Turn the vector by 90 degrees clockwise.
  * Add to the vector a certain vector C.



Operations could be performed in any order any number of times.

Can Gerald cope with the task?

Input

The first line contains integers x1 и y1 — the coordinates of the vector A ( - 108 ≤ x1, y1 ≤ 108). The second and the third line contain in the similar manner vectors B and C (their coordinates are integers; their absolute value does not exceed 108).

Output

Print "YES" (without the quotes) if it is possible to get vector B using the given operations. Otherwise print "NO" (without the quotes).

Examples

Input

0 0
1 1
0 1


Output

YES


Input

0 0
1 1
1 1


Output

YES


Input

0 0
1 1
2 2


Output

NO
#include <bits/stdc++.h>
const double eps = 1e-9;
const double pi = acos(-1);
using namespace std;
bool divis(const complex<long long>& a, const complex<long long>& b) {
  complex<long long> C(0, 0);
  C = a / b;
  return (b * C == a);
}
int main() {
  long long x, y;
  cin >> x >> y;
  complex<long long> A(x, y);
  cin >> x >> y;
  complex<long long> B(x, y);
  cin >> x >> y;
  complex<long long> C(x, y);
  complex<long long> I(0, 1);
  int ans = 0;
  if (C.real() == 0 && C.imag() == 0) {
    if (A == B)
      ans = 1;
    else if (-A == B)
      ans = 1;
    else if (B == A * I)
      ans = 1;
    else if (B == -A * I)
      ans = 1;
  } else {
    if (divis(B - A, C))
      ans = 1;
    else if (divis(B + A, C))
      ans = 1;
    else if (divis(B - A * I, C))
      ans = 1;
    else if (divis(B + A * I, C))
      ans = 1;
  }
  if (ans)
    cout << "YES\n";
  else
    cout << "NO\n";
  return 0;
}
