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>
using namespace std;
long long a1, a2, b1, b2, c1, c2, x, y;
long long aaabs(long long a) { return (a >= 0) ? a : -a; }
void solve() {
  cin >> a1 >> a2 >> b1 >> b2 >> c1 >> c2;
  if (c1 == 0 && c2 == 0) {
    if (a1 == b1 && a2 == b2) {
      cout << "YES";
      return;
    }
    if (a1 == -b1 && a2 == -b2) {
      cout << "YES";
      return;
    }
    if (a1 == b2 && a2 == -b1) {
      cout << "YES";
      return;
    }
    if (a1 == -b2 && a2 == b1) {
      cout << "YES";
      return;
    }
    cout << "NO";
    return;
  }
  x = b1 - a1;
  y = b2 - a2;
  if (aaabs(c2 * x - c1 * y) % (c1 * c1 + c2 * c2) == 0 &&
      aaabs(c1 * x + c2 * y) % (c1 * c1 + c2 * c2) == 0) {
    cout << "YES";
    return;
  }
  x = b1 + a1;
  y = b2 + a2;
  if (aaabs(c2 * x - c1 * y) % (c1 * c1 + c2 * c2) == 0 &&
      aaabs(c1 * x + c2 * y) % (c1 * c1 + c2 * c2) == 0) {
    cout << "YES";
    return;
  }
  x = b1 - a2;
  y = b2 + a1;
  if (aaabs(c2 * x - c1 * y) % (c1 * c1 + c2 * c2) == 0 &&
      aaabs(c1 * x + c2 * y) % (c1 * c1 + c2 * c2) == 0) {
    cout << "YES";
    return;
  }
  x = b1 + a2;
  y = b2 - a1;
  if (aaabs(c2 * x - c1 * y) % (c1 * c1 + c2 * c2) == 0 &&
      aaabs(c1 * x + c2 * y) % (c1 * c1 + c2 * c2) == 0) {
    cout << "YES";
    return;
  }
  cout << "NO";
}
int main() {
  solve();
  return 0;
}
