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 Ax, Ay, Bx, By, Cx, Cy;
bool solvable(long long A, long long B) {
  return (Cx * A + Cy * B) % (Cx * Cx + Cy * Cy) == 0 &&
         (-Cy * A + Cx * B) % (Cx * Cx + Cy * Cy) == 0;
}
bool possible() {
  cin >> Ax >> Ay >> Bx >> By >> Cx >> Cy;
  if (Cx == 0 && Cy == 0) {
    if (Bx == Ax && By == Ay) return true;
    if (Bx == Ay && By == -Ax) return true;
    if (Bx == -Ax && By == -Ay) return true;
    if (Bx == -Ay && By == Ax) return true;
    return false;
  }
  if (solvable(Bx - Ax, By - Ay)) return true;
  if (solvable(Bx - Ay, By + Ax)) return true;
  if (solvable(Bx + Ax, By + Ay)) return true;
  if (solvable(Bx + Ay, By - Ax)) return true;
  return false;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout << (possible() ? "YES" : "NO") << endl;
  cout << flush;
  return 0;
}
