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;
struct vect {
  long long x, y;
  vect() {}
  vect(long long a, long long b) {
    x = a;
    y = b;
  }
};
vect operator+(const vect& a, const vect& b) {
  return vect(a.x + b.x, a.y + b.y);
}
vect operator-(const vect& a, const vect& b) {
  return vect(a.x - b.x, a.y - b.y);
}
vect mul(const vect& a) { return vect(-a.y, a.x); }
bool solve(vect c, vect d) {
  if (c.x == 0) {
    if (c.y != 0) {
      if (d.x % c.y == 0 && d.y % c.y == 0) return true;
      return false;
    } else
      return d.x == 0 && d.y == 0;
  }
  long long p = d.y * c.x - c.y * d.x;
  long long q = c.x * c.x + c.y * c.y;
  if (p % q) return false;
  long long b = p / q;
  if ((d.x + b * c.y) % c.x) return false;
  return true;
}
int main() {
  vect a, b, c;
  cin >> a.x >> a.y >> b.x >> b.y >> c.x >> c.y;
  if (solve(c, b - a) || solve(c, b - mul(a)) || solve(c, b + a) ||
      solve(c, b + mul(a)))
    printf("YES");
  else
    printf("NO");
  return 0;
}
