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>
long long x[3], y[3], a1, a2, b1, b2;
int f = 0, i, a, b;
int zz() {
  long long t;
  a1 = x[1] - x[0], b1 = x[2], a2 = y[1] - y[0], b2 = y[2];
  t = x[0], x[0] = y[0], y[0] = -t;
  return 0;
}
int main() {
  for (i = 0; i < 3; i++) scanf("%I64d %I64d", &x[i], &y[i]);
  for (i = 0; i < 4; i++) {
    zz();
    if (b1 && b2) {
      b = (a1 * b1 + a2 * b2) / (b1 * b1 + b2 * b2), a = (b * b2 - a2) / b1;
      if (b * b1 + a * b2 == a1 && b * b2 - a * b1 == a2) f = 1;
    } else if (!b1 && !b2) {
      if (!a1 && !a2) f = 1;
    } else if (!b1) {
      if ((int)(a1 / b2) * b2 == a1 && (int)(a2 / b2) * b2 == a2) f = 1;
    } else if (!b2) {
      if ((int)(a1 / b1) * b1 == a1 && (int)(a2 / b1) * b1 == a2) f = 1;
    }
  }
  if (f)
    printf("YES\n");
  else
    printf("NO\n");
}
