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:33554432")
using namespace std;
const double PI = 2 * acos(0.0);
const double EPS = 1e-8;
const int INF = (1 << 30) - 1;
long long x0, yyyyy0, x1, yyyyy1, x, y;
bool can(long long x0, long long yyyyy0, long long x, long long y) {
  long long n = ((x0 * x + yyyyy0 * y) >= 0 ? (x0 * x + yyyyy0 * y)
                                            : -(x0 * x + yyyyy0 * y));
  ;
  long long d = x * x + y * y;
  if (d == 0) {
    return (x0 == 0 && yyyyy0 == 0);
  };
  if (n % d > 0) return false;
  n = ((x0 * y - yyyyy0 * x) >= 0 ? (x0 * y - yyyyy0 * x)
                                  : -(x0 * y - yyyyy0 * x));
  ;
  if (n % d > 0) return false;
  return true;
};
int main() {
  cin >> x0 >> yyyyy0 >> x1 >> yyyyy1 >> x >> y;
  for (int i = 0; i < 4; ++i) {
    if (can(x1 - x0, yyyyy1 - yyyyy0, x, y)) {
      cout << "YES";
      return 0;
    };
    long long tx, ty;
    tx = yyyyy1;
    ty = -x1;
    x1 = tx;
    yyyyy1 = ty;
  };
  cout << "NO";
  return 0;
};
