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 a1, a2, b1, b2, c1, c2;
struct Point {
  long long x, y;
  Point(long long x = 0, long long y = 0) : x(x), y(y) {}
};
Point ans[5];
int main() {
  scanf("%lld %lld %lld %lld %lld %lld", &a1, &a2, &b1, &b2, &c1, &c2);
  ans[1] = Point(b1 - a1, b2 - a2);
  ans[2] = Point(b1 - a2, b2 + a1);
  ans[3] = Point(b1 + a1, b2 + a2);
  ans[4] = Point(b1 + a2, b2 - a1);
  if (c1 == 0 && c2 == 0) {
    for (int i = 1; i <= 4; i++)
      if (ans[i].x == 0 && ans[i].y == 0) {
        printf("YES\n");
        return 0;
      }
    printf("NO\n");
    return 0;
  }
  for (int i = 1; i <= 4; i++) {
    long long t1, t2;
    if ((c1 * ans[i].x + c2 * ans[i].y) % (c1 * c1 + c2 * c2) != 0) continue;
    if ((c2 * ans[i].x - c1 * ans[i].y) % (c1 * c1 + c2 * c2) != 0) continue;
    printf("YES\n");
    return 0;
  }
  printf("NO\n");
  return 0;
}
