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 i, j, k, l, m, n, d, e, f, a, c, b, t, s, g, h;
map<long long, long long> y, x;
string p;
int main() {
  cin >> i >> j >> n >> m >> c >> d;
  k = c * c + d * d;
  if (k == 0) {
    if (i == n && j == m) t = 1;
    if (i == -n && j == -m) t = 1;
    if (j == n && -i == m) t = 1;
    if (-j == n && i == m) t = 1;
    if (t == 1)
      cout << "YES";
    else
      cout << "NO";
    return 0;
  }
  a = n - i;
  b = m - j;
  e = a * c + b * d;
  f = a * d - b * c;
  if (e % k == 0 && f % k == 0) t = 1;
  a = n + i;
  b = m + j;
  e = a * c + b * d;
  f = a * d - b * c;
  if (e % k == 0 && f % k == 0) t = 1;
  a = n + j;
  b = m - i;
  e = a * c + b * d;
  f = a * d - b * c;
  if (e % k == 0 && f % k == 0) t = 1;
  a = n - j;
  b = m + i;
  e = a * c + b * d;
  f = a * d - b * c;
  if (e % k == 0 && f % k == 0) t = 1;
  if (t == 1)
    cout << "YES";
  else
    cout << "NO";
}
