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>
const int magicconst = 73743071;
using namespace std;
long long a[2], x, y, b[2];
bool check(long long u, long long v) {
  long long det = x * x + y * y;
  long long q = x * v - u * y;
  long long w = u * x + y * v;
  return (q % det == 0 && w % det == 0);
}
int main() {
  scanf("%I64d%I64d%I64d%I64d%I64d%I64d", &a[0], &a[1], &b[0], &b[1], &x, &y);
  if (x == 0 && y == 0) {
    bool ans = 0;
    if (b[0] == a[0] && b[1] == a[1]) ans = 1;
    if (b[0] == -a[0] && b[1] == -a[1]) ans = 1;
    if (b[0] == a[1] && b[1] == -a[0]) ans = 1;
    if (b[0] == -a[1] && b[1] == a[0]) ans = 1;
    if (ans)
      printf("YES");
    else
      printf("NO");
  } else {
    bool ans = 0;
    if (check(b[0] - a[0], b[1] - a[1])) ans = 1;
    if (check(b[0] + a[0], b[1] + a[1])) ans = 1;
    if (check(b[0] - a[1], b[1] + a[0])) ans = 1;
    if (check(b[0] + a[1], b[1] - a[0])) ans = 1;
    if (ans)
      printf("YES");
    else
      printf("NO");
  }
  return 0;
}
