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>
#pragma comment(linker, "/STACK:16777216")
using namespace std;
struct point {
  long long x, y;
  void scan() { scanf("%lld%lld", &x, &y); }
} A, B, C;
bool can(long long a, long long b) {
  if (a == 0 && b == 0) return true;
  long long ch = a * C.y - b * C.x;
  long long zn = C.y * C.y + C.x * C.x;
  if (zn == 0) return false;
  if (ch % zn != 0) return false;
  long long y = ch / zn;
  ch = a - y * C.y;
  zn = C.x;
  if (zn == 0) {
    return y * C.y == a && b % C.y == 0;
  }
  if (ch % zn != 0) return false;
  return true;
}
bool pos() {
  return can(B.x - A.x, B.y - A.y) || can(B.y - A.x, -B.x - A.y) ||
         can(-B.x - A.x, -B.y - A.y) || can(-B.y - A.x, B.x - A.y);
}
int main() {
  A.scan();
  B.scan();
  C.scan();
  for (int i = 0; i < 4; ++i) {
    if (pos()) {
      printf("YES");
      return 0;
    }
    swap(C.x, C.y);
    C.x = -C.x;
  }
  printf("NO");
  return 0;
}
