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>
int main() {
  long long a1, a2, b1, b2, c1, c2, D;
  scanf("%lld%lld%lld%lld%lld%lld", &a1, &a2, &b1, &b2, &c1, &c2);
  if (c1 == 0 && c2 == 0) {
    if ((a1 == b1 && a2 == b2) || (a1 == -b1 && a2 == -b2) ||
        (a1 == b2 && a2 == -b1) || (a1 == -b2 && a2 == b1))
      printf("YES\n");
    else
      printf("NO\n");
    return 0;
  }
  D = c1 * c1 + c2 * c2;
  long long b1a1 = b1 + a1, b1_a1 = b1 - a1, b1a2 = b1 + a2, b1_a2 = b1 - a2,
            b2a2 = b2 + a2, b2_a2 = b2 - a2, b2a1 = b2 + a1, b2_a1 = b2 - a1;
  if (((((b1a1)*c1 + (b2a2)*c2) % D) || (((b1a1)*c2 - (b2a2)*c1) % D)) &&
      ((((b1_a1)*c1 + (b2_a2)*c2) % D) || (((b1_a1)*c2 - (b2_a2)*c1) % D)) &&
      ((((b1_a2)*c1 + (b2a1)*c2) % D) || (((b1_a2)*c2 - (b2a1)*c1) % D)) &&
      ((((b1a2)*c1 + (b2_a1)*c2) % D) || (((b1a2)*c2 - (b2_a1)*c1) % D)))
    printf("NO\n");
  else
    printf("YES\n");
  return 0;
}
