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;
int main() {
  cin >> ax >> ay >> bx >> by >> cx >> cy;
  long long dx = bx - ax, dy = by - ay, D = cx * cx + cy * cy;
  if (!D) {
    complex<double> a = complex<double>(ax, ay), b = complex<double>(bx, by),
                    I = complex<double>(0.0, 1.0);
    cout << (b == a || b == a * I || b == a * I * I || b == a * I * I * I
                 ? "YES"
                 : "NO");
    return 0;
  }
  if ((cx * dx + cy * dy) % D == 0 && (-cy * dx + cx * dy) % D == 0) {
    cout << "YES";
    return 0;
  }
  dx = bx + ay, dy = by - ax;
  if ((cx * dx + cy * dy) % D == 0 && (-cy * dx + cx * dy) % D == 0) {
    cout << "YES";
    return 0;
  }
  dx = bx + ax, dy = by + ay;
  if ((cx * dx + cy * dy) % D == 0 && (-cy * dx + cx * dy) % D == 0) {
    cout << "YES";
    return 0;
  }
  dx = bx - ay, dy = by + ax;
  if ((cx * dx + cy * dy) % D == 0 && (-cy * dx + cx * dy) % D == 0) {
    cout << "YES";
    return 0;
  }
  cout << "NO";
}
